# How do you find the axis of symmetry, and the maximum or minimum value of the function f(x)= 4x^2+40x+97?

May 20, 2016

Axis of symmetry is $x = - 5$ and minima at $\left(- 5 , - 3\right)$

#### Explanation:

For an equation of a parabola given by the equation

$y = a {x}^{2} + b x + c$, the axis of symmetry is a vertical line given by $x = - \frac{b}{2} a$

Hence, axis of symmetry for $y = 4 {x}^{2} + 40 x + 97$ is $x = - \frac{40}{2 \times 4} = - 5$

As the differential $\frac{\mathrm{dy}}{\mathrm{dx}} = 8 x + 40$ and this is zero at $8 x + 40 = 0$ or $x = - 5$. At this value $y = 4 \left(- 5\right) {x}^{2} + 40 \left(- 5\right) + 97 = 100 - 200 + 97 = - 3$

As second derivative (d^2y)/dx^2)=8 and is positive

hence we have a miniima at $\left(- 5 , - 3\right)$

graph{4x^2+40x+97 [-7.52, -2.52, -3.53, -1.03]}