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

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

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Answer 1 · 2018-06-16T11:40:39

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

Explanation:

$y = 4 x^{2} + 8 x + 7$ $y=4x^2+8x+7$

Take out the common factor
$y = 4 (x^{2} + 2 x) + 7$ $y=4(x^2+2x)+7$

Complete the square
$y = 4 (x^{2} + 2 x + 1) + 7 - 4 (1)$ $y=4(x^2+2x+1)+7-4(1)$

Simplify
$y = 4 {(x + 1)}^{2} + 3$ $y=4(x+1)^2+3$

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

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

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

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

$4 {(x + 1)}^{2} = - 3$ $4(x+1)^2=-3$

${(x + 1)}^{2} = - \frac{3}{4}$ $(x+1)^2=-3/4$

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 · 2018-06-16T11:55:00

Vertex is at $(- 1, 3)$ $(-1,3)$
y-intercept is at $(0, 7)$ $(0,7)$
(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:
$XXX y = m {(x - a)}^{2} + b$ $color(white)("XXX")y=color(green)m(x-color(red)a)^2+color(blue)b$
$XXX$ $color(white)("XXX")$ with vertex at $(a, b)$ $(color(red)a,color(blue)b)$

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

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

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

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

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

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

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

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

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

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

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

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

2 Answers

Explanation:

Explanation:

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