Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
1. Section 7.1
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Exercise 17 Page 321

The x-intercept is the point of intersection of the function with the x-axis. Similarly, the y-intercept is the point of intersection with the y-axis.

x-intercepts: - 52, 2
y-intercept: -10

Practice makes perfect

We want to find the x- and y-intercepts of a quadratic function. Let's begin with the x-intercept.

Finding the x-intercept

Think of the point where the graph of an equation crosses the x-axis. This is the x-intercept. The y-value of that ( x, y) coordinate pair is 0, so to find the x-intercept of the equation we should substitute 0 for y and solve for x.

y=2x^2+x-10
0=2x^2+x-10

We received a quadratic equation. To solve it, we can use the Quadratic Formula.

ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a Let's identify the values of a, b, and c in our case. 0=2x^2+x-10 ⇓ 2x^2+ 1x+( -10)=0 We can see that a= 2, b= 1, and c= -10. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a
x=- 1±sqrt(1^2-4( 2)( -10))/2( 2)
â–¼
Solve for x and Simplify
x=-1±sqrt(1-4(2)(-10))/2(2)
x=-1±sqrt(1-(-80))/2(2)
x=-1±sqrt(81)/4
x=-1±9/4

The solutions for this equation are x= -1±94.

x=-1±9/4
x_1=-1-9/4 x_2=-1+9/4
x_1=-10/4 x_2=8/4
x_1=-5/2 x_2=2

Using the Quadratic Formula, we found that the solutions of the given equation are x_1=- 52 and x_2=2. These are the x-intercepts.

Finding the y-intercept

Let's use the same concept to find the y-intercept. Consider the point where the graph of the equation crosses the y-axis. The x-value of the ( x, y) coordinate pair at the y-intercept is 0. Therefore, substituting 0 for x will give us the y-intercept.

y=2x^2+x-10
y=2( 0)^2+ 0-10
y=0+0-10
y=-10

A y-intercept of -10 means that the graph passes through the y-axis at the point (0, -10).