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

Jul 6, 2016

Vertex: (""(2,3))
$f \left(x\right)$-intercept: $15$
$x$-intercepts: (no Real intercepts); Complex intercepts $2 \pm i$

#### Explanation:

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

$f \left(x\right) = 3 {x}^{2} - 12 x + 15$

$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = 3 \left({x}^{2} - 4 x\right) + 15$

$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = 3 \left({x}^{2} - 4 x + 4\right) + 15 - 3 \times 4$

$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = \textcolor{g r e e n}{3} {\left(x - \textcolor{red}{2}\right)}^{2} + \textcolor{b l u e}{3}$
which is the vertex form for a parabola with vertex at $\left(\textcolor{red}{2} , \textcolor{b l u e}{3}\right)$

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The $f \left(x\right)$-intercept is the value of $f \left(x\right)$ when $x = 0$
Using the original form of the equation we can see that
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = 15$ when $x = 0$

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The $x$-intercept(s) is/are the value of $x$ when $f \left(x\right) = 0$.
Using the vertex form we derived
$\textcolor{w h i t e}{\text{XXX}} 0 = 3 {\left(x - 2\right)}^{2} + 3$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow {\left(x - 2\right)}^{2} + 1 = 0$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow {\left(x - 2\right)}^{2} = - 1$

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
$\textcolor{w h i t e}{\text{XXX}} x - 2 = \pm \sqrt{- 1} = \pm i$

$\textcolor{w h i t e}{\text{XXX}} \rightarrow x = 2 \pm i$

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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]}