Question

How do you find the vertex and the intercepts for `f(x)=3x^2-12x+15`?

Answer 1

Vertex: `(""(2,3))`
`f(x)`-intercept: `15`
`x`-intercepts: (no Real intercepts); Complex intercepts `2+-i`

Explanation:

To find the vertex convert the given equation into vertex form.

`f(x)=3x^2-12x+15`

`color(white)("XXX")f(x)=3(x^2-4x)+15`

`color(white)("XXX")f(x)=3(x^2-4x+4)+15-3xx4`

`color(white)("XXX")f(x)=color(green)(3)(x-color(red)(2))^2+color(blue)(3)`
which is the vertex form for a parabola with vertex at `(color(red)(2),color(blue)(3))`

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The `f(x)`-intercept is the value of `f(x)` when `x=0`
Using the original form of the equation we can see that
`color(white)("XXX")f(x)=15` when `x=0`

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The `x`-intercept(s) is/are the value of `x` when `f(x)=0`.
Using the vertex form we derived
`color(white)("XXX")0=3(x-2)^2+3`

`color(white)("XXX")rarr (x-2)^2+1=0`

`color(white)("XXX")rarr (x-2)^2=-1`

Since the square of any Real number is greater than or equal to zero
we can see that this equation has no Real solutions.

However if Complex solutions are permitted
`color(white)("XXX")x-2=+-sqrt(-1)=+-i`

`color(white)("XXX")rarr x=2+-i`

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For verification purposes, here is a graph of the original equation:

graph{3x^2-12x+15 [-21.36, 19.18, -0.08, 20.21]}