f(x) = a^2x +bx + c a ≠0
For a quadratic function to have a maximum value, the parameter a has to be negative. Furthermore, we can simplify this problem if we choose b = 0. With this decision, the maximum will occur at the y-axis. Since c represents the y-intercept, we just need to set c equal to the required maximum value.
Example Solution [0.8em]
a= - 2 b = c = 8 [0.8em]
f(x) = -2x^2 + ( )x + 8 [0.8em]
f(x) = -2 x^2 + 8
Notice that there are infinitely many equations satisfying the requirements. We just need to follow the conditions b=0, c=8, and a<0.
b Let's start from the standard quadratic form.
f(x) = a^2x +bx + c a ≠0
For a quadratic function to have a minimum value the parameter a has to be positive. Furthermore, we can simplify this problem if we choose b = 0. With this decision, the maximum will occur at the y-axis. Since c represents the y-intercept, we just need to set c equal to the required minimum value.
Example Solution [0.8em]
a= 3 b = c = - 4 [0.8em]
f(x) = 3x^2 + ( )x + (- 4) [0.8em]
f(x) = 3x^2 - 4
Notice that there are infinitely many equations satisfying these requirements. We just need to follow the conditions b=0, c=- 4, and a>0.
c Recall that we can find the x-coordinate of the vertex using the formula x = - b2a. Let's substitute x=-2, this will give us a condition for a and b.
The vertex will be at the x-coordinate x=-2 as long as b=4a. Therefore, there are infinitely many possible solutions. We can choose, for example, a=2 and b=8. Our equation, so far, is y= 2x^2 + 8x + c. We can now evaluate it at x=- 2 since we know the value of the function there is y=6, and solve for c.
