A parabola has x-intercepts at 3 and 7, a y-intercept at -21, and (5,4) for its vertex. Write the parabola’s equation

Sketch those points on a graph.

It is an upside down parabola (sheds water)
The axis of symmetry is x = 5
when x = 5, y = 4
so
for starters
y-4 = c (x-5)^2
That is the same both sides of x = 5
and y = 4 when x = 5
We need c
put in one of those other points
for example (0,-21)
-21 -4 = c (-5)^2
-25 = c (25)
c = -1
so
y-4 = -(x-5)^2
or
y = 4 - (x-5)^2

check with (3,0)
? = 4 -(-2)^2
? = 4 - 4
0 sure enough

Given: r1(3, 0), r2(7, 0). y-int.(0, -21). V(5, 4).

Y = a(x-h)^2 + k.
a(3 - 5)^2 + 4 = 0,
4a = -4,
a = -1.

Vertex form: Y = -1(x - 5)^2 + 4.

Y = -1(x^2 - 10x + 25) + 4,
Y = -x^2 + 10x - 21(Standard form).

To write the equation of a parabola given the x-intercepts, y-intercept, and vertex, we can use the standard form of the equation: y = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex.

First, let's find the values of h and k using the given information:
- The x-intercepts are at 3 and 7, so the parabola crosses the x-axis at these points. This means the vertex must lie on the axis of symmetry, which is the line equidistant from the x-intercepts. Therefore, the axis of symmetry is located at x = (3 + 7) / 2 = 5.
- The y-intercept is given as -21, which means the parabola crosses the y-axis at (0, -21). Since the vertex is halfway between the x-intercepts, we can determine that the vertex is also halfway between the y-intercept and the x-axis. Therefore, the y-coordinate of the vertex is (-21 + 0) / 2 = -10.5.

So, the vertex is (5, -10.5).

Now, substitute the values of h and k into the equation: y = a(x - 5)^2 - 10.5.

The only remaining unknown is the value of "a". To solve for "a", we can use the coordinates of another point on the parabola. In this case, we can use the given point (5, 4) on the parabola.

Substituting x = 5 and y = 4 into the equation, we get:
4 = a(5 - 5)^2 - 10.5
4 = a(0)^2 - 10.5
4 = 0 - 10.5
4 = -10.5

Since 4 does not equal -10.5, this means no value of "a" will satisfy this equation. Therefore, there is no parabola that satisfies the given conditions of having x-intercepts at 3 and 7, a y-intercept at -21, and a vertex at (5, 4).

In conclusion, there is no parabola that meets the given requirements.