Let's go through the two questions in your image step by step.

Question 1:

The function given is $y = -2x^2 - 4x + 4$.

To find the vertex of a quadratic function $y = ax^2 + bx + c$, we use the vertex formula: $x = \frac{-b}{2a}$ For the given function:

• $a = -2$
• $b = -4$

Now, let's calculate $x$: $x = \frac{-(-4)}{2(-2)} = \frac{4}{-4} = -1$

Now that we have the $x$-coordinate of the vertex, substitute $x = -1$ back into the original equation to find the $y$-coordinate: $y = -2(-1)^2 - 4(-1) + 4 = -2(1) + 4 + 4 = -2 + 8 = 6$

Thus, the vertex is $(-1, 6)$.

Because the leading coefficient $a = -2$ is negative, the parabola opens downwards, meaning the vertex is a maximum.

The correct answer is $(-1, 6)$; maximum.

Question 2:

The function given is $f(x) = x^2 + 3x - 10$.

Again, we use the vertex formula: $x = \frac{-b}{2a}$ For this function:

• $a = 1$
• $b = 3$

Let's calculate $x$: $x = \frac{-3}{2(1)} = \frac{-3}{2}$

Now substitute $x = -\frac{3}{2}$ back into the equation to find $y$: $f\left(-\frac{3}{2}\right) = \left(-\frac{3}{2}\right)^2 + 3\left(-\frac{3}{2}\right) - 10$