find the quadratic function in x that cuts the x axis at(-7/2,0) and (5/2,0)
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
To find the quadratic function that cuts the x-axis at the points (-7/2, 0) and (5/2, 0), we can start by using the standard form of a quadratic function:
f(x) = a(x - h)^2 + k
where (h, k) is the vertex of the parabola.
Since the parabola cuts the x-axis at (-7/2, 0) and (5/2, 0), the vertex will be the midpoint of these x-intercepts. The x-coordinate of the vertex is the average of the x-intercepts:
h = (-7/2 + 5/2)/2
h = -1/2
Since the parabola cuts the x-axis at these points, the y-coordinate of the vertex is zero:
k = 0
Now we can substitute the values of h and k into the standard form equation to get the final quadratic function:
f(x) = a(x - (-1/2))^2 + 0
f(x) = a(x + 1/2)^2
And that's the quadratic function that cuts the x-axis at (-7/2, 0) and (5/2, 0).