Find the [latex]y[/latex]-intercept of a quadratic function.
Find the real-number [latex]x[/latex]-intercepts, or roots of a quadratic function using factoring and the quadratic formula.
Use a graphing calculator to find the [latex]x[/latex]-intercepts and [latex]y[/latex]-intercept of a quadratic function.
When solving application problems involving quadratic functions, we often need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the [latex]y[/latex]-intercept of a quadratic by evaluating the function at an input of zero, and we find the [latex]x[/latex]-intercepts at locations where the output is zero. Notice that the number of [latex]x[/latex]-intercepts can vary depending upon the location of the graph.
Number of [latex]x[/latex]-intercepts of a parabola
Tip for success
We’ve built up several names for the values where a function crosses the horizontal axis:
horizontal intercepts or x-intercepts
zeros
roots
We tend to talk about intercepts with regard to a graph, zeros as solutions to an equation, and roots with regard to a function. Each term refers to the same value, though, for each quadratic function.
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Ex. [latex]f(x)=x^2-x-12[/latex]
The x-intercepts of the graph are located at [latex](-3, 0)[/latex] and [latex](4, 0)[/latex].
The zeros of the equation [latex]0 = x^2-x-12[/latex] are [latex]-3[/latex] and [latex]4[/latex].
The function[latex] f(x) = x^2 - x -12[/latex] has roots [latex]-3[/latex] and [latex]4[/latex].
How To: Given a quadratic function [latex]f\left(x\right)[/latex], find the y– and x-intercepts.
Evaluate [latex]f\left(0\right)[/latex] to find the [latex]y[/latex]-intercept.
Solve the quadratic equation [latex]f\left(x\right)=0[/latex] to find the [latex]x[/latex]-intercepts.
Example: Finding the y– and x-Intercepts of a Parabola
Find the [latex]y[/latex]– and [latex]x[/latex]-intercepts of the quadratic [latex]f\left(x\right)=3{x}^{2}+5x - 2[/latex].
Show Solution
We find the [latex]y[/latex]-intercept by evaluating [latex]f\left(0\right)[/latex].
So the [latex]y[/latex]-intercept is at [latex]\left(0,-2\right)[/latex].
For the [latex]x[/latex]-intercepts, or roots, we find all solutions of [latex]f\left(x\right)=0[/latex].
[latex]3{x}^{2}+5x - 2=0[/latex]
In this case, the quadratic can be factored easily, providing the simplest method for solution.
[latex]\left(3x - 1\right)\left(x+2\right)=0[/latex]
[latex]\begin{align}&3x - 1=0&x+2=0\end{align}[/latex]
[latex]x=\frac{1}{3}\hspace{8mm}[/latex] or [latex]\hspace{8mm}x=-2[/latex]
So the roots are at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex]\left(-2,0\right)[/latex].
Analysis of the Solution
By graphing the function, we can confirm that the graph crosses the [latex]y[/latex]-axis at [latex]\left(0,-2\right)[/latex]. We can also confirm that the graph crosses the [latex]x[/latex]-axis at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex]\left(-2,0\right)[/latex].
In the above example the quadratic was easily solved by factoring. Here is a video with more examples of finding the zeros or roots of a quadratic equation using factoring. https://youtu.be/zc2CpyRtjvY
Note: If you would like a quick review over factoring polynomials, then watch this video. https://youtu.be/U6FndtdgpcA
However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form (recall that this can also be referred to as vertex form).
How To: Given a quadratic function, find the x-intercepts by rewriting in standard form.
Substitute [latex]a[/latex] and [latex]b[/latex] into [latex]h=-\dfrac{b}{2a}[/latex].
Substitute [latex]x=h[/latex] into the general form of the quadratic function to find [latex]k[/latex].
Rewrite the quadratic in standard form using [latex]h[/latex] and [latex]k[/latex].
Solve for when the output of the function will be zero to find the [latex]x[/latex]–intercepts.
Example: Finding the Roots of a Parabola
Find the [latex]x[/latex]-intercepts of the quadratic function [latex]f\left(x\right)=2{x}^{2}+4x - 4[/latex].
Show Solution
We begin by solving for when the output will be zero.
[latex]0=2{x}^{2}+4x - 4[/latex]
Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.
The graph has [latex]x[/latex]–intercepts at [latex]\left(-1-\sqrt{3},0\right)[/latex] and [latex]\left(-1+\sqrt{3},0\right)[/latex].
Analysis of the Solution
We can check our work by graphing the given function on a graphing utility and observing the roots.
Try It
Using a graphing calculator or an online application like desmos, plot the function [latex]g\left(x\right)=-7+{x}^{2}-6x[/latex]. Find the [latex]x[/latex]-and [latex]y[/latex]-intercepts. Here is a video showing how to do this using a TI-84 graphing calculator.
When using desmos, type in your function. Then you can use your mouse to move around on the graph. If you click on a point on the graph, desmos will show you the coordinates for the point.
Find the x-and y-intercepts and list them as ordered pairs.
Vertex = [latex](3,-16)[/latex]
Show Solution
[latex]y[/latex]-intercept at [latex](0, -7)[/latex]
[latex]x[/latex]-intercept at [latex](-1,0)[/latex]
[latex]x[/latex]-intercept at [latex](7,0)[/latex]
tip for success
When attempting to identify the roots of a quadratic function, first look to see if it can be factored when the function value is set equal to zero, and solved by applying the zero-product principle. If so, this is often the quickest method. But most quadratic functions cannot be factored. In this case, you can either write it in vertex form or use the quadratic formula.
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1. The quadratic formula will reveal the x-intercepts of the graph of any quadratic equation.
3. The y-intercept, often called the initial value, of the function can be found by setting the input variable to 0.
Evaluate the function at zero to find the y-intercept: [latex]f(0) = y[/latex]
These three pieces of information are of particular value when solving applications of quadratic functions.
Example: Solving a Quadratic Equation with the Quadratic Formula
Solve [latex]{x}^{2}+x+2=0[/latex].
Show Solution
Let’s begin by writing the quadratic formula: [latex]x=\dfrac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex].
When applying the quadratic formula, we identify the coefficients [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex]. For the equation [latex]{x}^{2}+x+2=0[/latex], we have [latex]a=1[/latex], [latex]b=1[/latex], and [latex]c=2[/latex]. Substituting these values into the formula we have:
The solutions to the equation are [latex]x=\dfrac{-1+i\sqrt{7}}{2}[/latex] and [latex]x=\dfrac{-1-i\sqrt{7}}{2}[/latex] or [latex]x=-\dfrac{1}{2}+\dfrac{i\sqrt{7}}{2}[/latex] and [latex]x=-\dfrac{1}{2}-\dfrac{i\sqrt{7}}{2}[/latex].
Analysis of the Solution
This quadratic equation has only non-real solutions.
Example: Applying the Vertex and x-Intercepts of a Parabola
A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation [latex]H\left(t\right)=-16{t}^{2}+80t+40[/latex].
a. When does the ball reach the maximum height?
b. What is the maximum height of the ball?
c. When does the ball hit the ground?
Show Solution
a. The ball reaches the maximum height at the vertex of the parabola.
Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.
[latex]t=\dfrac{-80+\sqrt{8960}}{-32}\approx 5.458\hspace{3mm}[/latex] or [latex]\hspace{3mm}t=\dfrac{-80-\sqrt{8960}}{-32}\approx -0.458[/latex]
Since the domain starts at [latex]t=0[/latex] when the ball is thrown, the second answer is outside the reasonable domain of our model. The ball will hit the ground after about 5.46 seconds.
Try It
A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation [latex]H\left(t\right)=-16{t}^{2}+96t+112[/latex].
a. When does the rock reach the maximum height?
b. What is the maximum height of the rock?
c. When does the rock hit the ocean?
Show Solution
a. 3 seconds b. 256 feet c. 7 seconds
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Question ID 121416. Provided by: Lumen Learning. License: CC BY: Attribution. License Terms: MathAS Community License CC-BY + GPL
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Question ID 15809. Authored by: Sousa,James, mb Lippman,David. License: CC BY: Attribution. License Terms: MathAS Community License CC-BY + GPL