Two questions involving quadratics. Next to a diagram of a parabola with a ma point with the y inter

You know that a parabola is represented by a second degree equation so, if the parabola intercepts the $x$ axis ( i.e. has $y=0$) at $(-1,0)$ and $(p,0)$, the equation has the form:
$y=a(x+1)(x-p)$
Now you know that the parabola passes through the point $(=,p)$ , so, substituting these coordinates in the equation you can find $a$:
$p=a(-p)\Rightarrow a=-1$
and you have the equation $y=-x^2+(p-1)x+p$
For the question 3b):
Substitute $y=x+p$ in the equation of the parabola $y=-x^2+(p-1)x+p$ so you have a second degree equation $x+p=-x^2+(p-1)x+p$ whose solutions are the abscissas of the common points of the line and the parabola, so the parabola is tangent iff this equation has only one real solution, and this is done iff its discriminant is null. So, write out $\mathrm{\Delta }=0$ and you have an equation in the unknown $p$ whose solutions are the searched values.

For 3b), the line intersect the parabola at the point $(0,p)$. For this line to be tangent to the parabola, the derivative at $x=0$ must be equal to $1$.
$y^{\mathrm{\prime }}=(p-1)-2x$
$1=(p-1)-2\cdot 0⟹p=2$
It is the only value of $p$ which satisfy the condition