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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?

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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 $\geq 0$

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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

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

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

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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 $p\in[5,8]$.

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