Question

Rewrite the function in standard form, intercept form, find the vertex, find the y-intercept, and find the x-intercepts.

Answer 1

Answer

y = x^2 + 8x + 15 the standard form.

y +1 = (x + 4)(x + 4) intercept form.

Vertex =  (-4, -1)

y-intercept is (0, 15)

x-intercept are (-3, 0) and (-5, 0)

Step-by-step explanation:

The given equation is in vertex form.

y = (x + 4)^2 - 1

To write in standard form, we have to expand the above function.

Here we use the formula (a + b)^2 = a^2 + 2ab + b^2

y = x^2+  2*4*x + 4^2 - 1

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

y = x^2 + 8x + 15 the standard form.

y +1 = (x + 4)(x + 4) intercept form.

Vertex = (h, k) = (-4, -1)

y-intercept

To find the y-intercept, plug in x =0 in the above equation.

y = (0 + 4 )^2 -1

y = 16 - 1

y = 15

y-intercept is (0, 15)

x-intercept

To find the x-intercept plug in y =0 and find the values of x.

(x + 4)^2  -1 =0

(x + 4 )^2 =1

Taking the square root on both sides, we get

x + 4 = ±1

x-intercept are (-3, 0) and (-5, 0)

Hope this will helpful.

Thank you.

Answer 2

Answer:

Standard form - y=x^{2}+8x+15

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

The vertex is  is (-4,-1)

y intercept=(0,15)

X intercepts (-5,0) and (-3,0)

Step-by-step explanation:

The given quadratic equation is

$y=(x+4)^{2} -1\\y=x^{2}+8x+16-1\\y=x^{2}+8x+15$

and we know that the standard form of the quadratic equation is

$y=ax^{2}+bx+c$

so on comparing with it the above is the standard form only and also we can get the value of a (coefficient of $x^{2}$) and b( coefficient of x) and constant c so we get them as follows

a=1, b=8 c= 15 now ac=16 so the two numbers whose product is 15 and add is 8 are 5,3

so writing the equation in the Intercept form is $y=(x+5)(x+3)$

now lets find the vertex,now recall that all parabolas are symmetrical. This means that the axis of symmetry is halfway between the x−intercepts or their average.

axis of symmetry $=\frac{-5-3}{2}=\frac{-8}{2}=-4$

This is also the x−coordinate of the vertex. To find the y−coordinate, plug the x−value into either form of the quadratic equation. We will use Intercept form.

$y=(-4+5)(-4+3)=-1$

so, the vertex is (-4,-1).

now Vertex form is written as $y=a(x-h)^{2}+k$ , where (h,k) is the vertex and a is the same as in the other two forms.

for Y intercept put x=0 in the given vertex form

y=$(0+4)^{2}-1=16-1=15$

y intercept=(0,15)

for X intercept put y=0 in intercept form

$0=(x+5)(x+3)$

X intercepts (-5,0) and (-3,0)