Determine a quadratic function `f(x)=ax^2+bx+c` whose graph has a horizontal tangent at (-1,-8) and the graph passes through (2,19).
(1) A quadratic function is everywhere continuous and differentiable.
(2) If a graph has a horizontal tangent line at a point, then the point is a critical value so either the first derivative does not exist or the first derivative of the function is zero at the point. As quadratics are everywhere differentiable, `f'(-1)=0`
(3) Given `f(x)=ax^2+bx+c` then `f'(x)=2ax+b` .
Since `f'(-1)=0` we have `-2a+b=0=>b=2a`
(4) Rewrite the function as `f(x)=ax^2+2ax+c` . Now we use the fact that the function includes the points (-1,-8) and (2,19).
`4a+4a+c=19` `8a+c=19` ` `
Solving the system we get `9a=27=>a=3,b=2a=6,c=a-8=-5`
(5) The quadratic we seek is `f(x)=3x^2+6x-5`
** As a check we know that a parabola has a horizontal tangent at its vertex, so the vertex is (-1,-8). Writing the quadratic in vertex form yields:
Since the parabola goes through (2,19) we can substitue to solve for `a` :
Thus the quadratic is `f(x)=3(x+1)^2-8=3x^2+6x-5` as above.