Answer:
f(x) = 1/25x² - 1
Step-by-step explanation:
Given that:
The quadratic function f(x) = y = ax² + bx + c
Replace (x,y) = (5,0)
0 = a5² + b5 + c
0 = 25a + 5b + c ---- (1)
The differential eqaution;dt/dx = 2ax + b at (x,y) = (0, -1) it has minimum.
Thus, dy/dx = 0
2ax + b = 0
2a(0) + b = 0
0 + b = 0
b = 0 --- (2)
Now, replace (x,y) = (0, - 1) into equation (1)
Then;
-1 = 0 + 0 + c
c = -1
From equation (1)
0 = 25a + 5(b) + c
0 = 25a + 5(0) + c
c = - 25a
a = - c/25
a = -(-1)/25
a = 1/25
Therefore; the derived quadratic equation:
y = ax² + bx + c
y = 1/25x² + (0)(x) - 1
y = 1/25x² - 1
f(x) = 1/25x² - 1
The quadratic function that goes through (5,0) and has a local minimum at (0, -1) is [tex]f(x) =\frac{1}{25}x^2 -1[/tex]
The quadratic function is given as:
[tex]f(x) = ax^2 + bx +c[/tex]
It goes through point (5,0).
So, we have:
[tex]a(5)^2 + b(5) +c = 0[/tex]
[tex]25a + 5b +c = 0[/tex]
Also, we have:
[tex]f(x) = ax^2 + bx +c[/tex]
Differentiate
[tex]f'(x) = 2ax + b[/tex]
Set to 0
[tex]2ax + b = 0[/tex]
Substitute 0 for a
[tex]b = 0[/tex]
It has a local minimum at (0,-1).
So, we have:
[tex]a(0)^2 +b(0) + c = -1[/tex]
[tex]c = -1[/tex]
Recall that:
[tex]25a + 5b +c = 0[/tex]
This becomes
[tex]25a + 5(0) - 1 = 0[/tex]
[tex]25a- 1 = 0[/tex]
Collect like terms
[tex]25a = 1[/tex]
Solve for a
[tex]a = \frac{1}{25}[/tex]
Hence, the quadratic function is:
[tex]f(x) =\frac{1}{25}x^2 -1[/tex]
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