# How do you find the vertex and intercepts for f(x)=3(x-2)^2+1 ?

Dec 15, 2015

Vertex: (2,1)
x-intercepts: None
y-intercept: (0,13)

#### Explanation:

The quadratic given is already in vertex form, so finding the vertex is relatively simple:

$y = a {\left(x - h\right)}^{2} + k \to \text{Vertex: } \left(h , k\right)$

In this case:

$y = 3 {\left(x - 2\right)}^{2} + 1 \to \text{Vertex: } \left(2 , 1\right)$

To find the intercepts, we need to convert to standard form by evaluating the right hand side:

$3 {\left(x - 2\right)}^{2} + 1 = 3 \left({x}^{2} - 4 x + 4\right) + 1 = 3 {x}^{2} - 12 x + 13$

The y-intercept is simply $\left(0 , c\right)$ where $c$ is the constant term. In this case, $c = 13$ and the y-intercept is $\left(0 , 13\right)$.

We can use the quadratic formula to find the x-intercepts:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{12 \pm \sqrt{{12}^{2} - 4 \left(3\right) \left(13\right)}}{2 \left(3\right)}$
$x = \frac{12 \pm \sqrt{- 12}}{6}$

And here we reach a problem, because the numerator contains an imaginary number ($\sqrt{- 12}$), meaning that there are no x-intercepts and no real roots to this quadratic (though there are still 2 complex roots).