# Determine a quadratic function f(x)= ax^2 + bx + c if its graph passes through the point (2,19) and it has a horizontal tangent at (-1, -8) 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...

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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).

`a-2a+c=-8` `a-c=8`

`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:

`f(x)=a(x+1)^2-8`

Since the parabola goes through (2,19) we can substitue to solve for `a` :

`19=a(2+1)^2-8=>9a=27=>a=3`

Thus the quadratic is `f(x)=3(x+1)^2-8=3x^2+6x-5` as above.

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