# How do you find the intercept and vertex of y = (x − 3)(4x + 2)?

Oct 3, 2017

Solution Part 1 of 2

${x}_{\text{intercepts}} \to x = 3 \mathmr{and} x = - \frac{1}{2}$

${y}_{\text{intercept}} = - 6$

Vertex$\to \left(x , y\right) = \left(\frac{5}{4} , - \frac{49}{4}\right)$

#### Explanation:

$\textcolor{b l u e}{\text{Determine the x-intercepts}}$

x-intercept is at $y = 0$

Set $y = 0 = \left(x - 3\right) \left(4 x + 2\right)$

Consider the case $\left(x - 3\right) = 0 \implies x = 3$
Consider the case $\left(4 x + 2\right) = 0 \to 2 \left(2 x + 1\right) = 0 \implies x = - \frac{1}{2}$
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$\textcolor{b l u e}{\text{Determine the vertex}}$

If you multiply out the brackets you have $+ 4 {x}^{2} \ldots .$. This means that the graph is of form $\cup$ thus the vertex is a minimum.

${x}_{\text{vertex}}$ will be half way between the x-intecpts

${x}_{\text{vertex}} = \frac{3 + \left(- \frac{1}{2}\right)}{2} = \frac{2.5}{2} = 1.25 \to \frac{5}{4}$

Thus ${y}_{\text{vertex}} = \left(\frac{5}{4} - 3\right) \left(5 + 2\right) = - \frac{7}{4} \times 7 = - \frac{49}{4}$
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$\textcolor{b l u e}{\text{Determine the y-intercept}}$

Multiplying out the constants in the brackets we have

$\left(- 3\right) \times \left(+ 2\right) = - 6$

Thus ${y}_{\text{intercept}} = - 6$

Oct 3, 2017

Building 'completion of the square'

#### Explanation:

Notice that the section for this question is 'Vertex Form ....'

$y = \left(x - 3\right) \left(4 x + 2\right) = 4 {x}^{2} - 10 x - 6$

Set $y = 4 \left({x}^{2} - \frac{10}{4} x\right) - 6$

Halve the coefficient of $x$

$y = 4 \left({x}^{2} - \frac{10}{8} x\right) - 6$

Remove the $x$ from $- \frac{10}{8} x$ and move the squaring to outside the brackets

$y = 4 {\left(x - \frac{10}{8}\right)}^{2} - 6$ This will not produce the original equation $\textcolor{w h i t e}{\text{ddddddddddddddddd}}$as we have not includes a correction
$\textcolor{w h i t e}{\text{dddddddddddddddddd}}$process.

$y = \textcolor{red}{4} {\left(x \textcolor{g r e e n}{- \frac{10}{8}}\right)}^{2} + k - 6$ Now it will!

Note that $\textcolor{red}{4} \times \textcolor{g r e e n}{{\left(- \frac{10}{8}\right)}^{2}} + k = 0$

$\implies k = - \frac{25}{4}$

So $k - 6 \to - \frac{25}{4} - \frac{24}{4} = - \frac{49}{4}$

So the final format is:

$y = 4 {\left(x - \frac{5}{4}\right)}^{2} - \frac{49}{4}$