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

Jun 16, 2018

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

#### Explanation:

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

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

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

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

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

Therefore, the vertex is $\left(- 1 , 3\right)$

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

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

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

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

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 $\left(- 1 , 3\right)$
y-intercept is at $\left(0 , 7\right)$
(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:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
$\textcolor{w h i t e}{\text{XXX}}$with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Given
$\textcolor{w h i t e}{\text{XXX}} y = 4 {x}^{2} + 8 x + 7$

Extracting the $\textcolor{g r e e n}{m}$ factor
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{4} \left({x}^{2} + 2 x\right) + 7$

Completing the square:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{4} \left({x}^{2} + 2 x \textcolor{m a \ge n t a}{+ 1}\right) + 7 \textcolor{m a \ge n t a}{- \left(\textcolor{g r e e n}{4} \cdot 1\right)}$

Re-writing as a squared binomial and simplifying the constants
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{4} {\left(x + 1\right)}^{2} + \textcolor{b l u e}{3}$

color(white)("XXX")y=color(green)m(x-color(red)(""(-1)))^2+color(blue)3
which is the vertex form with vertex at $\left(\textcolor{red}{- 1} , \textcolor{b l u e}{3}\right)$

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

based on the original equation
$\textcolor{w h i t e}{\text{XXX")y=4x^2+8x+7color(white)("xxx}}$ with $x = 0$
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow y = 7$

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

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

from our derived vertex form:
$\textcolor{w h i t e}{\text{XXX")y=4(x-(-1))^2+3color(white)("xxx}}$ with $y = 0$
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow 4 {\left(x + 1\right)}^{2} = \textcolor{m a \ge n t a}{-} 3$
but
$\textcolor{w h i t e}{\text{XXX}}$for all Real values $4 {\left(x + 1\right)}^{2} > 0$
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow$ no value of $x$ satisfies this requirement.