# How do you find the vertex and intercepts for y =-3x^2 + 12x – 7?

##### 1 Answer
May 4, 2017

$\text{Vertex } \to \left(x , y\right) = \left(+ 2 , 5\right)$

${y}_{\text{intercept}} = - 7$

x_("intercepts")=2+-sqrt(5/3)" " Exact values

Approximate values ${x}_{\text{intercepts}}$ to 3 decimal places

$x \approx + 3.291$
$x \approx - 0.709$

#### Explanation:

See https://socratic.org/s/aEpUuXgB for a step by step example of method.

Given the standardised form of $y = a {x}^{2} + b x + c$

we have the vertex version of the same equation:

$y = a {\left(x + \frac{b}{2 a}\right)}^{2} + k + c$

Where $k = \left(- 1\right) \times a {\left(\frac{b}{2 a}\right)}^{2}$

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$\textcolor{b l u e}{\text{Determine vertex}}$

Write as $y = - 3 {\left(x + \frac{12}{- 6}\right)}^{2} + \left[\left(- 3\right) \times {\left(\frac{12}{- 6}\right)}^{2}\right] + 7$

$y = - 3 {\left(x - 2\right)}^{2} + 5$

$\textcolor{b l u e}{\text{Vertex } \to \left(x , y\right) = \left(+ 2 , 5\right)}$
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$\textcolor{b l u e}{\text{Determine y-intercepts}}$

Consider $y = - 3 {x}^{2} + 12 x - 7$

$\textcolor{b l u e}{\text{y-intercept } \to y = - 7}$

This occurs at $x = 0$
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$\textcolor{b l u e}{\text{Determine x-intercepts}}$

Note that we have $y = - 3 {x}^{2} + \ldots . .$

The coefficient of ${x}^{2}$ is negative so the graph is of general shape $\cap$

The vertex is above the x-axis so as of form $\cap$ there are x-intercepts

Set $y = 0$ giving:

$y = - 3 {\left(x - 2\right)}^{2} + 5 \text{ "->" } 0 = - 3 {\left(x - 2\right)}^{2} + 5$

Subtract 5 from both sides

$\text{ } - 5 = - 3 {\left(x - 2\right)}^{2}$

Divide both sides by -3

$\text{ } + \frac{5}{3} = {\left(x - 2\right)}^{2}$

Square root both sides

$\text{ } \pm \sqrt{\frac{5}{3}} = x - 2$

add 2 to both sides

$\text{ "x=2+-sqrt(5/3)" }$ Exact values

Approximate values to 3 decimal places

$x \approx + 3.291$
$x \approx - 0.709$