# How do you find the vertex of  f(x)= -3x^2-6x-7?

Jan 17, 2016

Vertex$\to \left(x , y\right) \to \left(- 1 , - 4\right)$

Solved using Calculus and by not using Calculus

#### Explanation:

As you are using $f \left(x\right)$ I am assuming you are at a higher level of Mathematics.
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$\textcolor{b r o w n}{\text{Using Calculus with shortcuts}}$

$\textcolor{b l u e}{\text{To find "x_("vertex}}$

Given: $f \left(x\right) = - 3 {x}^{2} - 6 x - 7$..............(1)

$f ' \left(x\right) = - 6 x - 6$

Equating to zero gives $\textcolor{b l u e}{{x}_{\text{vertex}} = - 1}$
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$\textcolor{b l u e}{\text{To find "y_("vertex}}$

Substitute $x = - 1$ into equation (1) giving

${y}_{\text{vertex}} = - 3 {\left(- 1\right)}^{2} - 6 \left(- 1\right) - 7$

$\textcolor{b l u e}{{y}_{\text{vertex}} = - 4}$
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Vertex$\to \left(x , y\right) \to \left(- 1 , - 4\right)$

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This is a maximum as $- 3 {x}^{2}$ being negative is indicative of an inverted U shape. Also the 2nd differential is negative which is also indicative of a maximum
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$\textcolor{b r o w n}{\text{NOT using Calculus}}$

$\textcolor{b l u e}{\text{I am going to show you a really cool trick!}}$

Write as:$y = \left(- 3 {x}^{2} - 6 x\right) - 7$

Factor the -3 out

$y = - 3 \left({x}^{2} + 2 x\right) - 7$............................(1)

Now consider the $+ 2 \text{ from } 2 x$ inside the bracket

Multiply this by $\left(- \frac{1}{2}\right)$

$\left(- \frac{1}{2}\right) \times \left(+ 2\right) = - 1$
This is the value you are after for $x$ so:

$\textcolor{b l u e}{{x}_{\text{vertex}} = - 1}$

Substitute this x value into the original equation to find the value of ${y}_{\text{vertex}}$