f(x) = x2 + 2x - 9 find minimum or maximum value then find coordinates of the maximum or minimum value
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`
Ah, the quadratic function! It always brings back memories of high school math. Anyway, let's find the minimum or maximum value of f(x) = x^2 + 2x - 9.
To find the minimum or maximum, we look at the coefficient of the x^2 term. In this case, it's positive, so we know that the graph opens upward, indicating a minimum point.
To find the x-coordinate of the minimum point, we use the formula x = -b / (2a), where a is the coefficient of x^2 and b is the coefficient of x. Plugging in the values, we have x = -2 / (2*1) = -1.
Now, to find the y-coordinate of the minimum point, we substitute x = -1 back into the equation f(x). So, f(-1) = (-1)^2 + 2(-1) - 9 = 1 - 2 - 9 = -10.
Therefore, the minimum value of the function is -10, and the coordinates of the minimum point are (-1, -10). Don't worry, it may be a bit negative, but you'll always find your way back up!
To find the minimum or maximum value of the function f(x) = x^2 + 2x - 9, we can start by finding the vertex of the parabola.
Step 1: The vertex of a quadratic function in the form f(x) = ax^2 + bx + c can be found using the formula x = -b / (2a).
In our case, a = 1, b = 2, and c = -9. Plugging these values into the formula, we get:
x = -2 / (2*1) = -2/2 = -1.
Step 2: To find the y-coordinate of the vertex, substitute the x-value (-1) into the original function f(x):
f(-1) = (-1)^2 + 2*(-1) - 9 = 1 - 2 - 9 = -10.
Therefore, the vertex of the parabola is (-1, -10).
Step 3: Since the coefficient of the x^2 term (a = 1) is positive, the parabola opens upwards, indicating a minimum value at the vertex.
Conclusion: The minimum value of the function f(x) = x^2 + 2x - 9 is -10, and the coordinates of the minimum are (-1, -10).
To find the minimum or maximum value of a quadratic function, we can use the concept of the vertex. The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the coordinates (h, k), where h = -b/2a and k = f(h).
For the given function f(x) = x^2 + 2x - 9, we can see that a = 1, b = 2, and c = -9.
First, let's find the x-coordinate of the vertex using h = -b/2a. Substituting the given values, we get h = -2/(2*1) = -1.
Next, we find the y-coordinate of the vertex by substituting h into the equation to get k = f(-1). Therefore, k = (-1)^2 + 2*(-1) - 9 = -1 - 2 - 9 = -12.
So, the vertex of the given function is (-1, -12).
Since the coefficient of the x^2 term is positive (a = 1 > 0), the parabola opens upwards. Thus, the vertex represents the minimum value of the function. Therefore, the minimum value of f(x) = x^2 + 2x - 9 is -12, and the coordinates of the minimum point are (-1, -12).