# How do you find the axis of symmetry, and the maximum or minimum value of the function f(x)=x ^2 + 3x-10?

Jan 23, 2016

Complete the square

#### Explanation:

$f \left(x\right) = {x}^{2} + 3 x - 10 \equiv \frac{{\left(2 x + 3\right)}^{2} - 49}{4}$

From here, we can see that

$f \left(- \frac{3}{2} - x\right) = \frac{4 {x}^{2} - 49}{4} = f \left(- \frac{3}{2} + x\right)$

Therefore the axis of symmetry is $x = - \frac{3}{2}$.

We also know that ${\left(2 x + 3\right)}^{2} \ge 0$. The minimum for $f$ corresponds to the value of $x$ which equality holds, i.e. ${\left(2 x + 3\right)}^{2} = 0$.

Solving it gives $x = - \frac{3}{2}$, which is unsurprising if you already know that the minimum/maximum of a parabola lies on its axis of symmetry.

$f \left(- \frac{3}{2}\right) = - \frac{49}{4}$ is the minimum.