f(x) = x^2 + 2x - 9
This is a quadratic equation. There are plenty of ways to get its maximum/minimum. One way is to graph and locate on the graph the point where max/min occurs.
Another is to solve for its vertex. We can transform the equation into the form
f(x) = a(x-h)^2 + k
where (h,k) is the vertex.
Therefore,
f(x) = x^2 + 2x - 9
f(x) = (x^2 + 2x) - 9
Completing the square:
f(x) = (x^2 + 2x + 1) - 9 - 1
f(x) = (x+1)^2 - 10
The vertex is therefore at (-1, -10).
Another way is to use derivatives. We get the derivative of the function with respect to x:
f(x) = x^2 + 2x - 9
f'(x) = 2x + 2
Then we equate it to zero (because at max/min, the slope of the tangent line is zero):
0 = 2x + 2
-2x = 2
x = -1
Substituting this back on the original function,
f(x) = x^2 + 2x - 9
f(1) = 1^2 + 2(-1) - 9
f(1) = 1 - 2 - 9
f(1) = -10
Thus, vertex is at (-1, -10).
By the way, this vertex is a minimum, because the numerical coefficient of x^2 in the function is positive.
hope this helps~ `u`
f(x) = x2 + 2x - 9 find minimum or maximum value then find coordinates of the maximum or minimum value
1 answer
1 answer