Question

1. the parabola with equation $y = a(x - 2)^2 + q$ passes through the points $(-2, 5)$ and $(4, -1)$. determine the coordinates of the vertex of the parabola.

Explanation:

Step1: Substitute (-2,5) into equation

Substitute $x=-2$, $y=5$ into $y=a(x-2)^2+q$:
$$5 = a(-2-2)^2 + q$$
$$5 = 16a + q$$

Step2: Substitute (4,-1) into equation

Substitute $x=4$, $y=-1$ into $y=a(x-2)^2+q$:
$$-1 = a(4-2)^2 + q$$
$$-1 = 4a + q$$

Step3: Solve for $a$ via elimination

Subtract the second equation from the first:
$$5 - (-1) = 16a + q - (4a + q)$$
$$6 = 12a$$
$$a = \frac{6}{12} = \frac{1}{2}$$

Step4: Solve for $q$

Substitute $a=\frac{1}{2}$ into $-1 = 4a + q$:
$$-1 = 4\times\frac{1}{2} + q$$
$$-1 = 2 + q$$
$$q = -3$$

Step5: Identify vertex coordinates

For vertex form $y=a(x-h)^2+k$, vertex is $(h,k)$. Here $h=2$, $k=q=-3$.

Answer:

$(2, -3)$