How do you find the vertex and the intercepts for f(x)=2x^2-12x+21?

Apr 17, 2017

$v = \left(6 , 21\right)$ and there are no interceptions.

Explanation:

The vertex is the minimum of the function, so if we set the standard notation $f \left(x\right) = a {x}^{2} + b x + c$ it is expressed as

${v}_{x} = - \frac{b}{2 a} = 6$ and ${v}_{y} = f \left({v}_{x}\right) = 21$

The intercepts are the roots of $f$, but $f$ has no real roots since

$\frac{\Delta}{4} = {\left(\frac{b}{2}\right)}^{2} - a c = 36 - 42 < 0$

Apr 17, 2017

Y-intercept: $\left(0 , 21\right)$
Vertex: $\left(3 , 3\right)$
No x intercepts

Explanation:

$f \left(x\right) = 2 {x}^{2} - 12 x + 21$

Y-intercept:
Find $f \left(0\right)$:
$f \left(0\right) = 2 {\left(0\right)}^{2} - 12 \left(0\right) + 21 = 21$

Vertex- Complete the square of the equation
$f \left(x\right) = 2 \left[{x}^{2} - 6 x + 10.5\right]$

$= 2 \left[{\left(x - 3\right)}^{2} + 1.5\right]$

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

Looking at the vertex form, the vertex is at $\left(3 , 3\right)$

X-intercepts:
Methods include using the quadratic formula, factoring, graphing, or completing the square/vertex form.
Using vertex form as found above:

$0 = f \left(x\right) = 2 {\left(x - 3\right)}^{2} + 3$
$- \frac{3}{2} = {\left(x - 3\right)}^{2}$
$x = \pm \sqrt{- \frac{3}{2}} + 3$

This is a square root of a negative number, meaning the function has no $x$ intercepts.