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

2 Answers
Jun 16, 2018

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

Jun 16, 2018

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.