Question

Find the x-intercepts of the parabola with
vertex (-3,-14) and y-intercept (0,13).
Write your answer in this form: (X1,Y1), (X2,42).
If necessary, round to the nearest hundredth.

Answer 1

Answer:

$(-5.16,0)$  and $(-0.84,0)$

Step-by-step explanation:

step 1

Find the equation of the quadratic equation

we know that

The equation of a vertical parabola into vertex form is equal to

$y=a(x-h)^{2}+k$

where

(h,k) is the vertex

a is a coefficient

we have that

(h,k)=(-3,-14)

substitute

$y=a(x+3)^{2}-14$

Remember that the y-intercept is the point (0,13)

substitute the value of x and y in the equation and fond the value of a

For x=0, y=13

$13=a(0+3)^{2}-14$

$13=9a-14$

$9a=27$

$a=3$

The equation is

$y=3(x+3)^{2}-14$

step 2

Find the x-intercepts

The x-intercepts are the values of x when the value of y is equal to zero

so

$0=3(x+3)^{2}-14$

$3(x+3)^{2}=14$

$(x+3)^{2}=14/3$

$x+3=(+/-)\sqrt{\frac{14}{3}}\\ \\x=-3(+/-)\sqrt{\frac{14}{3}}$

therefore

the x-intercepts are

$(-3-\sqrt{\frac{14}{3}},0)$ and  $(-3+\sqrt{\frac{14}{3}},0)$

or

$(-5.16,0)$  and $(-0.84,0)$

see the attached figure to better understand the problem