# What is the vertex, axis of symmetry, the maximum or minimum value, and the range of parabola f(x) = −4(x − 8)^2 + 3?

May 3, 2015

$f \left(x\right) = - 4 {\left(x - 8\right)}^{2} + 3$
is a standard quadratic in vertex form:
$f \left(x\right) = m {\left(x - a\right)}^{2} + b$
where $\left(a , b\right)$ is the vertex.

The fact that $m = - 4 < 0$ indicates that the parabola opens downward (the vertex is a maximum value)

The vertex is at $\left(8 , 3\right)$

Since it is a standard position parabola, the axis of symmetry is
$x = 8$

The maximum value is $3$

The Range of $f \left(x\right)$ is $\left(- \infty , + 3\right]$