Question

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

Answer 1

Axis of symmetry is `x=-5` and minima at `(-5,-3)`

Explanation:

For an equation of a parabola given by the equation

`y=ax^2+bx+c`, the axis of symmetry is a vertical line given by `x=-b/2a`

Hence, axis of symmetry for `y=4x^2+40x+97` is `x=-40/(2xx4)=-5`

As the differential `(dy)/(dx)=8x+40` and this is zero at `8x+40=0` or `x=-5`. At this value `y=4(-5)x^2+40(-5)+97=100-200+97=-3`

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

hence we have a miniima at `(-5,-3)`

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