Question

For the quadratic function (x+4)^2-1" align="absmiddle" class="latex-formula"> , do the following: a) rewrite the function in the standard form, b) rewrite the function in intercept form, c) find the vertex, d) find the y-intercept, e) find the x-intercepts.

Answer 1

Answer:

$y=x^2+8x+15$

Intercept form is y= (x+5)(x+3)

vertex is (-4,1)

y intercept is (0,15)

x intercepts are (-5,0) (-3,0)

Step-by-step explanation:

Given quadratic function

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

To rewrite it in standard form we need to expand (x+4)^2

$y=(x+4)(x+4)-1$

$y=x^2+8x+16-1$

$y=x^2+8x+15$

Intercept form is y=a(x-p)(x-q)

Factor x^2 + 8x + 15

product is 15 and sum is 8, 5*3= 15 and 5+3=8

(x+5)(x+3)

Intercept form is y= (x+5)(x+3)

Vertex form of equation is y=(x-h)^2+k

where (h,k) is the vertex

$y=(x+4)^2-1$, here h = -4 and k =1 so vertex is (-4,1)

To find y intercept plug in 0 for x

y= (x+5)(x+3)

y= (0+5)(0+3) = 15, so y intercept is (0,15)

to find x intercept plug in 0 for y

y= (x+5)(x+3)

0= (x+5)(x+3),             x+5 =0           and x+3=0

                                   x=-5              and x= -3

So x intercepts are (-5,0) (-3,0)