find the quadratic function in x that cuts the x axis at(-7/2,0) and (5/2,0)

Answer 1

To find the quadratic function that cuts the x-axis at (-7/2, 0) and (5/2, 0), we need to find the equation of the quadratic function in standard form: f(x) = ax^2 + bx + c.

Since both points have a y-coordinate of 0 when they intersect the x-axis, we can substitute these values into the equation to get two equations:

For (-7/2, 0):
0 = a(-7/2)^2 + b(-7/2) + c

For (5/2, 0):
0 = a(5/2)^2 + b(5/2) + c

Simplifying the equations, we have:
0 = 49a/4 - 7b/2 + c
0 = 25a/4 + 5b/2 + c

Now, we have a system of equations:

49a/4 - 7b/2 + c = 0
25a/4 + 5b/2 + c = 0

To solve this system, we can multiply the second equation by -2 to eliminate c:

49a/4 - 7b/2 + c = 0
-25a/2 - 5b + 2c = 0

Adding the two equations together, we obtain:

24a/4 - 12b/2 = 0
6a - 6b = 0

Simplifying further:

6a = 6b
a = b

Now, we know that a = b. Using this fact, let's assign a value to a, say a = 1:

6(1) = 6b
b = 1

So, a = 1 and b = 1. Plugging these values back into the equations:

49(1)/4 - 7(1)/2 + c = 0
c = -49/4 + 7/2

Simplifying further:

c = -49/4 + 14/4
c = -35/4

Therefore, the quadratic function that cuts the x-axis at (-7/2, 0) and (5/2, 0) is:

f(x) = x^2 + x - 35/4