Quadratics Question

Question

Find the x-intercepts and the x-coordinate of the vertex of the graph representing each equation. If you get stuck, try graphing.

1. y = x^2 - 25
2. y = x^2 + 16

Answer 1

First, think about what you know about x-intercepts. They are the points when y = 0, so to find them, you can plug y into the equation and then solve for x.

0 = x² - 25

Remember that a negative number, when squared, gives a positive answer, so in this case, the x intercept is both 5 and -5. This make sense because this is a parabola with its vertex below the x axis, so it should have two x-intercepts.

To find the vertex, think about the vertex form of the equation, which is y = a (x - h)² +k, where (h,k) is the vertex. In this example, there is no h term and the k is -25, so the vertex is (0,25).

For the second parabola, you would follow the same format. However, think about the x-intercepts. Can you take a square root of a negative number? What does that tell you about how many x-intercepts there are for the second graph?

Answer 2

1) y=x^2-25

To find the x-intercepts, we will set y to be 0 and solve for x:

0=x^2-25,

25=x^2

x= 5 or -5

x-intercepts: (-5, 0) and (5, 0)

For the vertex of these parabolas, x will be (-b/2a)

Imagine the quadratic ax^2+bx+c...well, our a is 1, our b is 0 (since there's no x^1 term), and our c is -25.

Since b is 0, we won't bother with the denominator since that makes x of the vertex 0.

Now plugging that in for x into the parabola: y=0^2-25, thus the vertex is (0, -25)

2) y=x^2+16

We will simply repeat the process:

0=x^2+16

-16=x^2

There is no real solution for x here, so there are no x-intercepts.

You can also plot this parabola on a graph and see that it never intercepts the x-axis.

We can still find the vertex,

x=0 (since there is no x^1 term)

y=0^2+16

vertex=(0, 16)