Relations, Functions and Graphs II

Setting $x = 0$ leaves only the constant term, so the curve crosses the $y$-axis at 10.

The coefficient of $x^2$ is $-1$, which is negative, so the curve opens downwards.

Half of 8 is 4, so $x^2 + 8x = (x+4)^2 - 16$. Then $-16 + 3 = -13$.

Set $y = 0$ and factorise: $(x-4)(x+2) = 0$. Two numbers multiplying to $-8$ and adding to $-2$ are $-4$ and $2$.

The bracket is zero when $x = 5$, giving the smallest value of $y$, which is 3. The axis of symmetry runs vertically through the turning point.

Completing the square gives $(x+1)^2 + 4$. A square is never negative, so $y$ is always at least 4 and cannot reach zero.