How do you know how many solutions a parabola graph has?

Two points: If the vertex of a U-shaped parabola lies below the x-axis or the the vertex of an upside down U-shaped parabola lies above the x-axis, then the parabola will cross the x-axis exactly twice . Thus, in this case there are two solutions.

How do you find the answer to a parabola?

To solve problems on parabolas the general equation of the parabola is used, it has the general form y = ax 2 + bx + c (vertex form y = a(x - h) 2 + k) where, (h,k) = vertex of the parabola.

What is the 1 3 5 rule for graphing parabolas?

The step pattern of the graph of f(x) = x2 is 1, 3, 5, .... This means that as you move away from the vertex in the x-direction, the graph goes up by 1, then 3, then 5, etc . All quadratic functions have the same step pattern, multiplied by the “a” value.

What are the features of a parabola 10.3 answers?

The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum (Figure 10.3.

What is the rule for graphing a parabola?

To draw a parabola graph, we have to first find the vertex for the given equation. This can be done by using x=-b/2a and y = f(-b/2a). Plotting the graph, when the quadratic equation is given in the form of f(x) = a(x-h) 2 + k, where (h, k) is the vertex of the parabola, is its vertex form .

How many points are needed to draw a parabola?

It is a fundamental fact that two points uniquely determine a line. That is, given two distinct points, there is one and only one line that passes through these points. How many (distinct) non-collinear points in the plane are required to determine a parabola? The answer is three.

What is the max of a parabola?

One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value.

How many solutions does the parabola have?

If the discriminant is positive, there are two distinct real solutions (the parabola intersects the x-axis at two distinct points). If the discriminant is zero, there is exactly one real solution (the parabola touches the x-axis at one point, known as a double root).

How to determine the number of solutions?

If we can solve the equation and get something like x=b where b is a specific number, then we have one solution . If we end up with a statement that's always false, like 3=5, then there's no solution. If we end up with a statement that's always true, like 5=5, then there are infinite solutions..

How to find how many real solutions a graph has?

If the graph of a quadratic function crosses the x-axis at two points, then the equation has two real rational solutions . These solutions are also called x-intercepts or roots. If it touches the x-axis at one point, it has one real rational solution.

How do you know if a parabola has one solution?

The only way the quadratic equation/parabola can have one solution is if the parabola intersects with the x-axis in one place . That means that the vertex of the parabola sits on the x-axis. The quickest way of determining this is to focus on the portion of the quadratic equation {eq}b^2 - 4ac {/eq}.

9.5: Graphing Parabolas (2024)

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Learning Objectives

Graph a parabola.
Find the intercepts and vertex of a parabola.
Find the vertex of a parabola by completing the square.

The Graph of a Quadratic Equation

We know that any linear equation with two variables can be written in the form $y=mx+b$ and that its graph is a line. In this section, we will see that any quadratic equation of the form $y=ax^{2}+bx+c$ has a curved graph called a parabola.

Two points determine any line. However, since a parabola is curved, we should find more than two points. In this text, we will determine at least five points as a means to produce an acceptable sketch. To begin, we graph our first parabola by plotting points. Given a quadratic equation of the form $y=ax^{2}+bx+c$, x is the independent variable and y is the dependent variable. Choose some values for x and then determine the corresponding y-values. Then plot the points and sketch the graph.

Example $\PageIndex{1}$

Graph by plotting points:

$y=x^{2}-2x-3$

Solution:

In this example, choose the x-values {−2, −1, 0, 1, 2, 3, 4} and calculate the corresponding y-values.

Plot these points and determine the shape of the graph.

Answer:

When graphing, we want to include certain special points in the graph. The y-intercept is the point where the graph intersects the y-axis. The x-intercepts are the points where the graph intersects the x-axis. The vertex is the point that defines the minimum or maximum of the graph. Lastly, the line of symmetry(also called the axis of symmetry) is the vertical line through the vertex, about which the parabola is symmetric.

For any parabola, we will find the vertex and y-intercept. In addition, if the x-intercepts exist, then we will want to determine those as well. Guessing at the x-values of these special points is not practical; therefore, we will develop techniques that will facilitate finding them. Many of these techniques will be used extensively as we progress in our study of algebra. Given a quadratic equation of the form $y=ax^{2}+bx+c$, find the y-intercept by setting x=0 and solving. In general, $y=a(0)^{2}+b(0)+c=c$, and we have

\(\color{Cerulean}{y-intercept}

\[(0,c]\)

Next, recall that the x-intercepts, if they exist, can be found by setting y=0. Doing this, we have $0=a^{2}+bx+c$, which has general solutions given by the quadratic formula, $x=\frac{−b±\sqrt{b^{2}−4ac}}{2a}$. Therefore, the x-intercepts have this general form:

$\color{Cerulean}{x-intercepts}$

Using the fact that a parabola is symmetric, we can determine the vertical line of symmetry using the x-intercepts. To do this, we find the x-value midway between the x-intercepts by taking an average as follows:

Therefore, the line of symmetry is the vertical line:

$\color{Cerulean}{Line\:of\:symmetry}$

\[x=-\frac{b}{2 a}\]

We can use the line of symmetry to find the x-value of the vertex. The steps for graphing a parabola are outlined in the following example.

Example $\PageIndex{2}$

Graph:

Solution

Step 1: Determine the y-intercept. To do this, set x = 0 and solve for y.

The y-intercept is (0, 3).

Step 2: Determine the x-intercepts. To do this, set y = 0 and solve for x.

$\begin{array}{rlr}{x+3} & {=0} & {\text { or }} & {x-1=0} \\ {x} & {=-3} & {} & {x=1}\end{array}$

Here when y = 0, we obtain two solutions. There are two x-intercepts, (−3, 0) and (1, 0).

Step 3: Determine the vertex. One way to do this is to use the equation for the line of symmetry, $x=\frac{-b}{2 a}$, to find the x-value of the vertex. In this example, a = −1 and b = −2:

Substitute −1 into the original equation to find the corresponding y-value.

The vertex is (−1, 4).

Step 4: Determine extra points so that we have at least five points to plot. In this example, one other point will suffice. Choose x = −2 and find the corresponding y-value.

Our fifth point is (-2,3).

Step 5: Plot the points and sketch the graph. To recap, the points that we have found are

Table $\PageIndex{1}$
y-intercept:	(0,3)
x-intercept:	(-3,0) and (1,0)
Vertex:	(-1,4)
Extra point:	(-2,3)

Answer:

The parabola opens downward. In general, use the leading coefficient to determine whether the parabola opens upward or downward. If the leading coefficient is negative, as in the previous example, then the parabola opens downward. If the leading coefficient is positive, then the parabola opens upward.

All quadratic equations of the form $y=ax^{2}+bx+c$ have parabolic graphs with y-intercept (0, c). However, not all parabolas have x intercepts.

Example $\PageIndex{3}$

Graph:

$y=2x^{2}+4x+5$

Solution:

Because the leading coefficient 2 is positive, note that the parabola opens upward. Here c = 5 and the y-intercept is (0, 5). To find the x-intercepts, set y = 0.

$\begin{array}{l}{y=2 x^{2}+4 x+5} \\ {0=2 x^{2}+4 x+5}\end{array}$

Table \(\PageIndex{2}\)
y-intercept:	(0,5)
x-intercepts:	None
Vertex:	(-1,3)
Extra points	(-3,11), (-2,5), (1,11)

Table \(\PageIndex{3}\)
y-intercept:	(0, -18)
x-intercepts:	(3, 0)
Vertex:	(3, 0)
Extra points:	(1, -8), (5, -8), (6, -18)

Table \(\PageIndex{4}\)
y-intercept:	(0, -1)
x-intercepts:	\((1-\sqrt{2}, 0)\) and \((1+\sqrt{2}, 0)\)
Vertex:	\((1, -2)\)
Extra point:	\((2, -1)\)

Finding the Maximum and Minimum

It is often useful to find the maximum and/or minimum values of functions that model real-life applications. To find these important values given a quadratic function, we use the vertex. If the leading coefficient a is positive, then the parabola opens upward and there will be a minimum y-value. If the leading coefficient a is negative, then the parabola opens downward and there will be a maximum y-value.

Example $\PageIndex{6}$

Determine the maximum or minimum:

$y=−4x^{2}+24x−35$

Solution:

Since a = −4, we know that the parabola opens downward and there will be a maximum y-value. To find it, we first find the x-value of the vertex.

$\begin{aligned} x &=-\frac{b}{2 a} \qquad\quad\color{Cerulean}{x-value\:of\:the\:vertex.} \\ &=-\frac{24}{2(-4)}\quad\:\color{Cerulean}{Substitute\:a=-4\:and\:b=24.} \\ &=-\frac{24}{-8} \qquad\:\: \color{Cerulean}{Simplify.} \\ &=3 \end{aligned}$

The x-value of the vertex is 3. Substitute this value into the original equation to find the corresponding y-value.

The vertex is (3, 1). Therefore, the maximum y-value is 1, which occurs when x = 3, as illustrated below:

Note

The graph is not required to answer this question.

Answer:

The maximum is 1.

Example $\PageIndex{7}$

Determine the maximum or minimum:

$y=4x^{2}−32x+62$

Solution:

Since a = +4, the parabola opens upward and there is a minimum y-value. Begin by finding the x-value of the vertex.

$\begin{aligned} x &=-\frac{b}{2 a} \\ &=-\color{black}{\frac{\color{OliveGreen}{-32}}{\color{black}{2}(\color{OliveGreen}{4}\color{black}{)}}}\qquad\color{Cerulean}{Substitute\:a=4\:and\:b=-32.} \\ &=-\frac{-32}{8}\qquad\color{Cerulean}{Simplify.} \\ &=4 \end{aligned}$

Substitute x = 4 into the original equation to find the corresponding y-value.

The vertex is (4, −2). Therefore, the minimum y-value of −2 occurs when x = 4, as illustrated below:

Answer:

The minimum is -2.

Exercise $\PageIndex{2}$

Determine the maximum or minimum:

$y=(x-3)^{2}-9$

Answer: The minimum is −9.

A parabola, opening upward or downward (as opposed to sideways), defines a function and extends indefinitely to the right and left as indicated by the arrows. Therefore, the domain (the set of x-values) consists of all real numbers. However, the range (the set of y-values) is bounded by the y-value of the vertex.

Finding the Vertex by Completing the Square

In this section, we demonstrate an alternate approach for finding the vertex. Any quadratic equation $y=ax^{2}+bx+c$ can be rewritten in the form

\[y=a(x-h)^{2}+k\]

In this form, the vertex is (h, k).

Example $\PageIndex{10}$

Determine the vertex:

$y=-4(x-3)^{2}+1$

Solution:

When the equation is in this form, we can read the vertex directly from the equation.

$\begin{array}{l}{y=\:a\:(\:x-h)^{2}+k} \\ \color{Cerulean}{\qquad\qquad\quad\downarrow\quad\:\:\downarrow} \\ {y=-4(x-3)^{2}+1}\end{array}$

Here h=3 and k=1.

Answer:

The vertex is (3, 1).

Example $\PageIndex{11}$

Determine the vertex:

$y=2(x+3)^{2}-2$

Solution:

Rewrite the equation as follows before determining h and k.

$\begin{array}{l}{y=\:a\:(\:x\:-h)^{2}\:\:\:\:+\:\:\:k} \\ \color{Cerulean}{\qquad\qquad\quad\:\downarrow\qquad\quad\downarrow} \\ {y=2(x-(-3))^{2}+(-2)}\end{array}$

Here h=-3 and k=-2.

Answer:

The vertex is (-3,-2).

Often the equation is not given in this form. To obtain this form, complete the square.

Example $\PageIndex{12}$

Rewrite in $y=a(x−h)^{2}+k$ form and determine the vertex: $y=x^{2}+4x+9$.

Solution:

Begin by making room for the constant term that completes the square.

$\begin{aligned} y &=x^{2}+4 x+9 \\ &=x^{2}+4 x+\underline\quad+9-\underline\quad\end{aligned}$

The idea is to add and subtract the value that completes the square, $\frac{b^{2}}{2}$, and then factor. In this case, add and subtract $\frac{4^{2}}{2} = 2^{2} = 4$.

$\begin{aligned} y &=x^{2}+4 x+9 \qquad\quad\:\:\:\color{Cerulean}{Add\:and\:subtract\:4.} \\ &=\underbrace{x^{2}+4 x\color{Cerulean}{+4}}_{\text { factor }}+9\color{Cerulean}{-4} \quad\color{Cerulean}{Factor.} \\ &=(x+2)(x+2)+5 \\ &=(x+2)^{2}+5 \end{aligned}$

Adding and subtracting the same value within an expression does not change it. Doing so is equivalent to adding 0. Once the equation is in this form, we can easily determine the vertex.

$\begin{array}{c}{y=a(x-h)^{2}\:+\:\:k} \\ \color{Cerulean}{\qquad\quad\quad\:\:\downarrow\qquad\downarrow} \\ {y=(x-(-2))^{2}+5}\end{array}$

Here h=-2 and k=5.

Answer:

The vertex is (-2, 5).

If there is a leading coefficient other than 1, then we must first factor out the leading coefficient from the first two terms of the trinomial.

Example $\PageIndex{13}$

Rewrite in $y=a(x−h)^{2}+k$ form and determine the vertex: $y=2x^{2}−4x+8$.

Solution:

Since a = 2, factor this out of the first two terms in order to complete the square. Leave room inside the parentheses to add a constant term.

Now use −2 to determine the value that completes the square. In this case, $\frac{(-2)^{2}}{2}=\((-1)^{2}=1$. Add and subtract 1 and factor as follows:

In this form, we can easily determine the vertex.

$\begin{array}{l}{y=a(x-h)^{2}+k} \\ \color{Cerulean}{\qquad\qquad\:\downarrow\quad\:\:\:\downarrow}\\ {y=2(x-1)^{2} \:+6}\end{array}$

Here h=1 and k=6.

Answer:

The vertex is (1, 6).

Exercise $\PageIndex{3}$

Rewrite in $y=a(x-h)^{2}+k$ form and determine the vertex:

$y=-2x^{2}-12x+3$.

Answer: $y=-2(x+3)^{2}+21$; vertex: $(-3, 21)$

Key Takeaways

The graph of any quadratic equation $y=ax^{2}+bx+c$, where a, b, and c are real numbers and a≠0, is called a parabola.
When graphing parabolas, find the vertex and y-intercept. If the x-intercepts exist, find those as well. Also, be sure to find ordered pair solutions on either side of the line of symmetry, $x=\frac{-b^{2}}{a}$.
Use the leading coefficient, a, to determine if a parabola opens upward or downward. If a is positive, then it opens upward. If a is negative, then it opens downward.
The vertex of any parabola has an x-value equal to $x=\frac{-b^{2}}{a}$. After finding the x-value of the vertex, substitute it into the original equation to find the corresponding y-value. This y-value is a maximum if the parabola opens downward, and it is a minimum if the parabola opens upward.
The domain of a parabola opening upward or downward consists of all real numbers. The range is bounded by the y-value of the vertex.
An alternate approach to finding the vertex is to rewrite the quadratic equation in the form $y=a(x−h)^{2}+k$. When in this form, the vertex is (h, k) and can be read directly from the equation. To obtain this form, take $y=ax^{2}+bx+c$ and complete the square.

Exercise $\PageIndex{4}$ the graph of quadratic equations

Does the parabola open upward or downward? Explain.

$y=x^{2}−9x+20$
$y=x^{2}−12x+32$
$y=−2x^{2}+5x+12$
$y=−6x^{2}+13x−6$
$y=64−x^{2}$
$y=−3x+9x^{2}$

Answer

1. Upward

3. Downward

5. Downward

Exercise $\PageIndex{5}$ the graph of quadratic equations

Determine the x- and y-intercepts.

$y=x^{2}+4x−12$
$y=x^{2}−13x+12$
$y=2x^{2}+5x−3$
$y=3x^{2}−4x−4$
$y=−5x^{2}−3x+2$
$y=−6x^{2}+11x−4$
$y=4x^{2}−25$
$y=9x^{2}−49$
$y=x^{2}−x+1$
$y=5x^{2}+15x$

Answer

1. x-intercepts: $(−6, 0), (2, 0)$; y-intercept: $(0, −12)$

3. x-intercepts: $(−3, 0), (\frac{1}{2}, 0)$; y-intercept: $(0, −3)$

5. x-intercepts: $(−1, 0), (\frac{2}{5}, 0)$; y-intercept: $(0, 2)$

7. x-intercepts: $(−\frac{5}{2}, 0), (\frac{5}{2}, 0)$; y-intercept: $(0, −25)$

9. x-intercepts: none; y-intercept: $(0, 1)$

Exercise $\PageIndex{6}$ the graph of quadratic equations

Find the vertex and the line of symmetry.

$y=−x^{2}+10x−34$
$y=−x^{2}−6x+1$
$y=−4x^{2}+12x−7$
$y=−9x^{2}+6x+2$
$y=4x^{2}−1$
$y=x^{2}−16$

Answer

1. Vertex: $(5, −9)$; line of symmetry: $x=5$

3. Vertex: $(\frac{3}{2}, 2)$; line of symmetry: $x= \frac{3}{2}$

5. Vertex: $(0, −1)$; line of symmetry: $x=0$

Exercise $\PageIndex{7}$ the graph of quadratic equations

Graph. Find the vertex and the y-intercept. In addition, find the x-intercepts if they exist.

$y=x^{2}−2x−8$
$y=x^{2}−4x−5$
$y=−x^{2}+4x+12$
$y=−x^{2}−2x+15$
$y=x^{2}−10x$
$y=x^{2}+8x$
$y=x^{2}−9$
$y=x^{2}−25$
$y=1−x^{2}$
$y=4−x^{2}$
$y=x^{2}−2x+1$
$y=x^{2}+4x+4$
$y=−4x^{2}+12x−9$
$y=−4x^{2}−4x+3$
$y=x^{2}−2$
$y=x^{2}−3$
$y=−4x^{2}+4x−3$
$y=4x^{2}+4x+3$
$y=x^{2}−2x−2$
$y=x^{2}−6x+6$
$y=−2x^{2}+6x−3$
$y=−4x^{2}+4x+1$
$y=x^{2}+3x+4$
$y=−x^{2}+3x−4$
$y=−2x^{2}+3$
$y=−2x^{2}−1$
$y=2x^{2}+4x−3$
$y=3x^{2}+2x−2$

Answer

11.

13.

15.

17.

19.

21.

23.

25.

27.

Exercise $\PageIndex{8}$ maximum or minimum

Determine the maximum or minimum y-value.

$y=−x^{2}−6x+1$
$y=−x^{2}−4x+8$
$y=25x^{2}−10x+5$
$y=16x^{2}−24x+7$
\(y=−x^{2}
$y=1−9x^{2}$
$y=20x−10x^{2}$
$y=12x+4x^{2}$
$y=3x^{2}−4x−2$
$y=6x^{2}−8x+5$

Answer

1. Maximum: $y = 10$

3. Minimum: $y = 4$

5. Maximum: $y = 0$

7. Maximum: $y = 10$

9. Minimum: $y = −\frac{10}{3}$

Exercise $\PageIndex{9}$ maximum or minimum

Given the following quadratic functions, determine the domain and range.

$f(x)=3x^{2}+30x+50$
$f(x)=5x^{2}−10x+1$
$g(x)=−2x^{2}+4x+1$
$g(x)=−7x^{2}−14x−9$
The height in feet reached by a baseball tossed upward at a speed of 48 feet/second from the ground is given by the function $h(t)=−16t^{2}+48t$, where t represents time in seconds. What is the baseball’s maximum height and how long will it take to attain that height?
The height of a projectile launched straight up from a mound is given by the function $h(t)=−16t^{2}+96t+4$, where t represents seconds after launch. What is the maximum height?
The profit in dollars generated by producing and selling x custom lamps is given by the function $P(x)=−10x^{2}+800x−12,000$. What is the maximum profit?
The revenue in dollars generated from selling a particular item is modeled by the formula $R(x)=100x−0.0025x^{2}$, where x represents the number of units sold. What number of units must be sold to maximize revenue?
The average number of hits to a radio station website is modeled by the formula $f(x)=450t^{2}−3,600t+8,000$, where t represents the number of hours since 8:00 a.m. At what hour of the day is the number of hits to the website at a minimum?
The value in dollars of a new car is modeled by the formula $V(t)=125t^{2}−3,000t+22,000$, where t represents the number of years since it was purchased. Determine the minimum value of the car.
The daily production costs in dollars of a textile manufacturing company producing custom uniforms is modeled by the formula $C(x)=0.02x^{2}−20x+10,000$, where x represents the number of uniforms produced.
1. How many uniforms should be produced to minimize the daily production costs?
2. What is the minimum daily production cost?
The area of a certain rectangular pen is given by the formula $A=14w−w^{2}$, where w represents the width in feet. Determine the width that produces the maximum area.

Answer

1. Domain: R; range: $[−25,∞)$

3. Domain: R; range: $(−∞,3]$

5. The maximum height of 36 feet occurs after 1.5 seconds.

7. $4,000

9. 12:00 p.m.

11. a. 500 uniforms; b. $5,000

Exercise $\PageIndex{10}$ vertex by completing the square

Determine the vertex.

$y=−(x−5)^{2}+3$
$y=−2(x−1)^{2}+7$
$y=5(x+1)^{2}+6$
$y=3(x+4)^{2}+10$
$y=−5(x+8)^{2}−1$
$y=(x+2)^{2}−5$

Answer

1. $(5, 3)$

3. $(-1, 6)$

5. $(-8, -1)$

Exercise $\PageIndex{11}$ vertex by completing the square

Rewrite in $y=a(x−h)^{2}+k$ form and determine the vertex.

$y=x^{2}−14x+24$
$y=x^{2}−12x+40$
$y=x^{2}+4x−12$
$y=x^{2}+6x−1$
$y=2x^{2}−12x−3$
$y=3x^{2}−6x+5$
$y=−x^{2}+16x+17$
$y=−x^{2}+10x$

Answer

1. $y=(x−7)^{2}−25$; vertex: $(7, −25)$

3. $y=(x+2)^{2}−16$; vertex: $(−2, −16)$

5. $y=2(x−3)^{2}−21$; vertex: $(3, −21)$

7. $y=−(x−8)^{2}+81$; vertex: $(8, 81)$

Exercise $\PageIndex{12}$ vertex by completing the square

Graph.

$y=x^{2}−1$
$y=x^{2}+1$
$y=(x−1)^{2}$
$y=(x+1)^{2}$
$y=(x−4)^{2}−9$
$y=(x−1)^{2}−4$
$y=−2(x+1)^{2}+8$
$y=−3(x+2)^{2}+12$
$y=−5(x−1)^{2}$
$y=−(x+2)^{2}$
$y=−4(x−1)^{2}−2$
$y=9(x+1)^{2}+2$
$y=(x+5)^{2}−15$
$y=2(x−5)^{2}−3$
$y=−2(x−4)^{2}+22$
$y=2(x+3)^{2}−13$

Answer

11.

13.

15.

Exercise $\PageIndex{13}$ discussion board

Write down your plan for graphing a parabola on an exam. What will you be looking for and how will you present your answer? Share your plan on the discussion board.
Why is any parabola that opens upward or downward a function? Explain to a classmate how to determine the domain and range.

Answer: 1. Answers may vary

9.5: Graphing Parabolas (2024)

The Graph of a Quadratic Equation

Finding the Maximum and Minimum

Finding the Vertex by Completing the Square

Key Takeaways

FAQs

How do you know how many solutions a parabola graph has? ›

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