Question

graph a parabola whose x-intercepts are at x = -3 and x = 5 and whose minimum value is y = -4.

Explanation:

Step1: Find vertex x-coordinate

The x-coordinate of the vertex (minimum point) is the midpoint of the x-intercepts:
$$x = \frac{-3 + 5}{2} = 1$$
So the vertex is $(1, -4)$.

Step2: Write intercept form of parabola

The intercept form is $y = a(x - p)(x - q)$, where $p=-3$, $q=5$:
$$y = a(x + 3)(x - 5)$$

Step3: Solve for coefficient $a$

Substitute vertex $(1, -4)$ into the equation:
$$-4 = a(1 + 3)(1 - 5)$$
$$-4 = a(4)(-4)$$
$$-4 = -16a$$
$$a = \frac{-4}{-16} = \frac{1}{4}$$

Step4: Write final equation

Substitute $a=\frac{1}{4}$ back into intercept form:
$$y = \frac{1}{4}(x + 3)(x - 5)$$
(To graph, plot the x-intercepts $(-3,0)$, $(5,0)$, the vertex $(1,-4)$, and draw a symmetric parabola opening upwards.)

Answer:

The parabola has equation $y = \frac{1}{4}(x + 3)(x - 5)$, with key points: x-intercepts at $(-3, 0)$ and $(5, 0)$, vertex (minimum) at $(1, -4)$. The graph is an upward-opening parabola passing through these points.