Answer:
Therefore, the equation for the parabola is y = 5(x - 1/2)^2 - 405/4.
Explanation:
To write an equation for a parabola that passes through the x-intercepts (-4,0) and (5,0) and the point (3,-70), we can use the standard form equation of a parabola: y = a(x - h)^2 + k.
1. Start by finding the values of h and k, which represent the coordinates of the vertex. The x-coordinate of the vertex is the average of the x-intercepts, (-4 + 5)/2 = 1/2, and the y-coordinate is the value of y at the vertex. We can substitute the x-coordinate into the equation to find the y-coordinate:
y = a(x - 1/2)^2 + k
0 = a((-4) - 1/2)^2 + k
0 = a((-8/2) - 1/2)^2 + k
0 = a(-9/2)^2 + k
0 = a(81/4) + k
2. Next, substitute the coordinates of the point (3,-70) into the equation to solve for a:
-70 = a(3 - 1/2)^2 + k
-70 = a(5/2)^2 + k
-70 = a(25/4) + k
3. Now we have a system of equations:
0 = a(81/4) + k
-70 = a(25/4) + k
4. We can solve this system of equations to find the values of a and k.
5. Subtract the second equation from the first to eliminate k:
0 - (-70) = a(81/4) + k - (a(25/4) + k)
70 = a(81/4) - a(25/4)
70 = (81a - 25a)/4
70 = 56a/4
70 = 14a
a = 70/14
a = 5
6. Substitute the value of a back into one of the equations to solve for k:
0 = 5(81/4) + k
0 = 405/4 + k
-405/4 = k
7. Finally, substitute the values of a and k back into the equation y = a(x - h)^2 + k:
y = 5(x - 1/2)^2 - 405/4