# Find Possible Values of p So That Parabola Is On or Above Line

The question is:

For what real values of $$p$$ will the graph of the parabola $$y=x^2-2px+p+1$$ be on or above that of the line $$y=-12x+5$$?

Therefore, the $$y$$ value of the vertex of the parabola must be greater than or equal to the $$y$$ value of the line for corresponding values of $$x$$

An attempt to translate that would be:

$$-12x+5 = -p^2+p+1$$

With "$$-p^2+p+1$$" being the $$y$$ value of the vertex, as mentioned. However, this attempt doesn't really seem to open up any further steps.

What would be a more proper approach/solution for this problem?

Notice that here it is enough to find an extremum of the function $$f(x)=(x^2-2px+p+1)-(-12x+5)$$ Find why this is a minimum, and then you can get an equation that gives you the required conditions on $$p$$ after you look at the cases where this minimum is $$\geq 0$$

You must have

$$(x^2-2px+p+1)-(-12x+5)\ge0$$

and by completing the square,

$$(x-p+6)^2-(p-6)^2+p+1-5\ge0.$$

This will be true for all $$x$$ when

$$-p^2+13p-40\ge0.$$

$$p\in[5,8].$$

The parabola is on or above the line if and only if the intersection between them has no more than a point. So, consider the system$$\left\{\begin{array}{l}y=x^2-2px+p+1\\y=-12x+5.\end{array}\right.$$This leads you to the quadratic equation$$-12x+5=x^2-2px+p+1\tag1$$and you're after the values of $$p$$ for which the discriminant is smaller than or equal to $$0$$. So, since $$(1)\iff x^2+(12-2p)x+p-4=0$$, the discriminant is$$(12-2p)^2-4(p-4)=4(p^2-13p+40),$$which happens to be smaller than or equal to $$0$$ if and only if $$p\in[5,8]$$.