Question

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

Answer 1

See explanation

Explanation:

As the coefficient of `x^2` is negative the graph is of general shape `nn`

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

`color(blue)("Determine the y-intercept")`

`color(green)("y-intercept is at the same value as the constant ie "y=-3)`

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`color(blue)("Determine the vertex")`

Write as `-1/6(x^2-6x)-3`

`color(green)(x_("vertex")=(-1/2)xx(-6) = +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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`color(blue)("Determine the x-intercepts")`

As the graph is of form `nn` and `y_("vertex")=-1.5`

The curve doe NOT cross the x-axis.

So there is no solution to `" "-1/6x^2+x-3=0`
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`color(blue)("Determine the axis of symmetry")`

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

`color(green)("Axis of symmetry "-> x=3)`