I'm having some troubles with these. Thanks in advance.
If a>0 find the minimum value. If a<0 find the maximum value.
1. y=-x²+2x+5
2. y=-2x²-3x+4
These are functions for parabolas, so
f(x)=ax²+bx+c when a does not equal 0
In both your examples, the "a", which is the coefficient of the x^2 term, is negative.
1. dy/dx = -2x + 2, and that equals zero at a maximum or minimum. So the maximum would occur at x=1.
d2y/dx^2 = -2. Since that is always negative, x=1 is a point of maximum value.
Do #2 similarly.
Thank-you, but I really don't understand what you mean by dy/dx. I'm feeling terribly stupid, but nothing like that is in my book.
Find Minimum or maximum value of f(x)=2x^2 + 3x -9
No problem at all! I apologize for the confusion. Let me explain it in a simpler way.
To find the minimum or maximum value of a quadratic function, you can use a concept called "vertex". The vertex represents the highest or lowest point on the parabola.
In order to find the vertex of a quadratic function in the form f(x) = ax^2 + bx + c, you can use the formula: x = -b / (2a).
For your function, f(x) = 2x^2 + 3x - 9, we can identify that a = 2, b = 3, and c = -9.
To find the x-coordinate of the vertex, substitute the values of a and b into the formula x = -b / (2a):
x = -(3) / (2 * 2) = -3/4
Now, substitute this x-coordinate into the original function to find the y-coordinate:
f(-3/4) = 2(-3/4)^2 + 3(-3/4) - 9
= 2(9/16) - 9/4 - 9
= 9/8 - 9/4 - 9
= 9/8 - 18/8 - 9
= -27/8 - 18/8
= -45/8
Therefore, the vertex of the function f(x) = 2x^2 + 3x - 9 is (-3/4, -45/8), where the x-coordinate represents the value of x that minimizes or maximizes the function, and the y-coordinate represents the corresponding minimum or maximum value.
In this case, since the coefficient of the x^2 term (a) is positive (2 > 0), the function opens upwards, and the vertex (-3/4, -45/8) represents the minimum value of the function.
If you have any further questions or need clarifications, please feel free to ask!