I cannot for the life of me remember how to do this- Find the vertex of the graphs of the functions:
Function #1: y=(x-4)(x+2)
AND
Function #2: y=2x^2-4x+1
Can someone help me get it PLz!
OK - to find the vertex of these functions you must have them in quadratic form where the first coefficient in front of the x^2 is a, the linear term is b, and the constant is c. You then do -b/2a. For the first one, you must FOIL it out to get x^2-2x-8. b is -2 and a is 1. So, the x-coordinate of the vertex is -(-2)/2(1) which is 1. Then plug it back in for x so y=(1-4)(1+2), y=-9. The vertex is (1,-9). Do the same for the second function. You should end up with (1,1); if you don't get this please post again.
To find the vertex of a quadratic function, you need to have the function in standard form, which is written as y = ax^2 + bx + c.
For the first function, y = (x-4)(x+2), we can expand it using the distributive property: y = x^2 + 2x - 4x - 8. Combining like terms, we get y = x^2 - 2x - 8.
Now, we can see that a (coefficient of x^2) is 1, b (coefficient of x) is -2, and c (constant term) is -8.
To find the x-coordinate of the vertex, you use the formula -b/2a. Plugging in the values, we have -(-2)/2(1) = 2/2 = 1.
So, the x-coordinate of the vertex is 1.
Next, we substitute this value back into the function to get the y-coordinate. y = (1-4)(1+2) = (-3)(3) = -9.
Therefore, the vertex of the first function is (1, -9).
For the second function, y = 2x^2 - 4x + 1, we can see that a is 2, b is -4, and c is 1.
Using the same formula, -b/2a, we get -(-4)/2(2) = 4/4 = 1.
So, the x-coordinate of the vertex is 1.
Next, we substitute this value back into the function to get the y-coordinate. y = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1.
Therefore, the vertex of the second function is (1, -1).