Question

How do you find the vertex and intercepts for `y = 4x^2 + 8x + 7`?

Answer 1

Therefore, the vertex is `(-1,3)` and the y-intercept is `(0,7)`

Explanation:

`y=4x^2+8x+7`

Take out the common factor
`y=4(x^2+2x)+7`

Complete the square
`y=4(x^2+2x+1)+7-4(1)`

Simplify
`y=4(x+1)^2+3`

which is in the form `y=(x-h)^2+k` where `(h,k)` is the vertex

Therefore, the vertex is `(-1,3)`

For y-intercept, sub `x=0`
`y=0+0+7`
`y=7`
`(0,7)`

For x-intercept, sub `y=0`
`4(x+1)^2+3=0`

`4(x+1)^2=-3`

`(x+1)^2=-3/4`

Therefore, no solution as any number squared is greater than or equal to 0.

graph{4x^2+8x+7 [-10, 10, -5, 5]}

Above is the parabola

Answer 2

Vertex is at `(-1,3)`
y-intercept is at `(0,7)`
(there is no x-intercept)

Explanation:

The easiest way (in this case) to find the vertex is to convert the given equation into vertex form:
`color(white)("XXX")y=color(green)m(x-color(red)a)^2+color(blue)b`
`color(white)("XXX")`with vertex at `(color(red)a,color(blue)b)`

Given
`color(white)("XXX")y=4x^2+8x+7`

Extracting the `color(green)m` factor
`color(white)("XXX")y=color(green)4(x^2+2x)+7`

Completing the square:
`color(white)("XXX")y=color(green)4(x^2+2xcolor(magenta)(+1))+7color(magenta)(-(color(green)4 * 1))`

Re-writing as a squared binomial and simplifying the constants
`color(white)("XXX")y=color(green)4(x+1)^2+color(blue)(3)`

Adjusting the sign inside the squared binomial to match the vertex form requirement
`color(white)("XXX")y=color(green)m(x-color(red)(""(-1)))^2+color(blue)3`
which is the vertex form with vertex at `(color(red)(-1),color(blue)3)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The y-intercept is the value of `y` when `x=0`

based on the original equation
`color(white)("XXX")y=4x^2+8x+7color(white)("xxx")` with `x=0`
`color(white)("XXX")rArr y=7`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The x-intercept is the value of `x` when `y=0`

from our derived vertex form:
`color(white)("XXX")y=4(x-(-1))^2+3color(white)("xxx")` with `y=0`
`color(white)("XXX")rArr 4(x+1)^2=color(magenta)-3`
but
`color(white)("XXX")`for all Real values `4(x+1)^2 > 0`
`color(white)("XXX")rArr` no value of `x` satisfies this requirement.