# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function f(x)=-3x^2+x-5?

Feb 26, 2018

Axis of symmetry is at $x = \frac{1}{6}$, the maximum value at $- \frac{59}{12}$ and minimum value at $- \infty$.

#### Explanation:

The axis of symmetry is basically the $x$-coordinate of the vertex. The $x$- coordinate of a parabola is given by:

$- \frac{b}{2 a}$, and here:

$a = - 3$ and $b = 1$. Inputting:

${A}_{s} = - \frac{b}{2 a}$

${A}_{s} = - \frac{1}{2 \cdot - 3}$

${A}_{s} = - \frac{1}{-} 6$

${A}_{s} = \frac{1}{6}$

Since the leading coefficient of the equation is negative, the graph opens downward. Therefore, the maximum value is the $y$ coordinate of the vertex. We have:

$y = - 3 {x}^{2} + x - 5$, and the vertex has $x = \frac{1}{6}$, so:

$y = - 3 {\left(\frac{1}{6}\right)}^{2} + \frac{1}{6} - 5$

$y = - 3 \left(\frac{1}{36}\right) + \frac{1}{6} - 5$

$y = - \frac{1}{12} + \frac{1}{6} - 5$

$y = - \frac{59}{12}$

The minimum value is so $- \infty$, as the parabola keeps going down.

Graphing such, we can confirm:

graph{-3x^2+x-5 [-9.63, 10.37, -10.64, -0.64]}

Feb 26, 2018

General shape $\cap$ thus a maximum

${y}_{\text{intercept}} = - 5$

"No "x_("intercepts")

Vertex $\to \left(x , y\right) = \left(\frac{1}{6} , - \frac{59}{12}\right)$

#### Explanation:

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

Write as $\textcolor{g r e e n}{y = - 3 \left({x}^{2} \textcolor{red}{- \frac{1}{3}} x\right) - 5}$
The above is the start of completing the square

$\textcolor{w h i t e}{\text{d}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{m a \ge n t a}{\text{The axis of symmetry is at}}$
${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(\textcolor{red}{- \frac{1}{3}}\right) = \textcolor{m a \ge n t a}{+ \frac{1}{6}}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{m a \ge n t a}{y \text{ intercept}}$
Consider the form $y = a {x}^{2} + b x + c$

${y}_{\text{intercept}} = c = - 5$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{m a \ge n t a}{\text{General shape of the graph}}$

As the ${x}^{2}$ term is negative the general shape is $\cap$

Thus the vertex is a maximum.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{m a \ge n t a}{\text{Determine the vertex}}$

Found above that ${x}_{\text{vertex}} = \frac{1}{6}$

color(green)(y=-3color(red)(x)^2+color(red)(x)-5 color(white)("ddd")->color(white)("ddd")y_("vertex")=-3(color(red)(1/6) )^2+color(red)(1/6)-5

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{dddddddddddddddddd")->color(white)("dd")y_("vertex}} = - \frac{59}{12}}$

As the graph is of form $\cap \mathmr{and} {y}_{\text{vertex}} = - \frac{59}{12}$ then the vertex is below the x-axis so the is NO x-intercepts.

Vertex $\to \left(x , y\right) = \left(\frac{1}{6} , - \frac{59}{12}\right)$

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$\textcolor{b l u e}{\text{Foot note}}$

They would be expecting you to solve for the vertex by completing the square.