# How do you graph the quadratic function and identify the vertex and axis of symmetry for y=-1/6x^2+x-3?

Feb 27, 2017

See explanation

#### Explanation:

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

There is no stipulation in the question about how you are to determine the information so I chose the following method

$\textcolor{b l u e}{\text{Determine the y-intercept}}$

$\textcolor{g r e e n}{\text{y-intercept is at the same value as the constant ie } y = - 3}$

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

Write as $- \frac{1}{6} \left({x}^{2} - 6 x\right) - 3$

$\textcolor{g r e e n}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(- 6\right) = + 3}$

Substitute $x = 3$ and we have

color(green)(y_("vertex") =-1/6( 3^2-(6xx3))-3" "=" "-1.5)

color(green)("Vertex "->(x,y)=(3,-3/2)
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$\textcolor{b l u e}{\text{Determine the x-intercepts}}$

As the graph is of form $\cap$ and ${y}_{\text{vertex}} = - 1.5$

The curve doe NOT cross the x-axis.

So there is no solution to $\text{ } - \frac{1}{6} {x}^{2} + x - 3 = 0$
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$\textcolor{b l u e}{\text{Determine the axis of symmetry}}$

The axis of symmetry passes through the vertex so it has the x value as the vertex.

$\textcolor{g r e e n}{\text{Axis of symmetry } \to x = 3}$