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.