Find the quadratic function f(x) = ax + bx+c that goes through (5,0) and has a local minimum at (0, -1). f(x)=

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 required function is .

step-by-step explanation:

from the given table it is noticed that the graph passing through (-3,7), (0,4), (3,1) and (8,-4).

it is a linear function because the value of y is decrease by one as the value of x increase by 1.

the slope of the function is

the slope of the function is -1 and the y-intercept is 4.

the slope intercept form is

where, m is slope and b is y-intercept.

therefore the required function is . the graph of this function is shown below.