F(x)=1/4 (x+7)^2+7
Looking for the vertex and the max or min for the above..? f(1/2)=19/2, did I find the vertex?
(6x^2-5) ( x^7-6) using FOIL. I came up with 6x^9-5x^7-36x^2+30 please tell me I'm right if not what the heck did I do?
Thanks...
6x^9 -36x^2 -35x^7 + 30 is what you should have gotten second equation using foil
oh wait correction i read it wrong you have it right sorry
Let's start by finding the vertex of the function f(x) = (1/4)(x+7)^2 + 7.
To find the vertex of the quadratic function f(x) = ax^2 + bx + c, you can use the formula:
x = -b/2a
In our case, a = 1/4 and b = 7. Plugging these values into the formula, we get:
x = -(7) / (2 * (1/4))
x = -28
To find the y-coordinate of the vertex, you can substitute this x-value back into the original function:
f(-28) = (1/4)((-28)+7)^2 + 7
f(-28) = (1/4)(-21)^2 + 7
f(-28) = (1/4)(441) + 7
f(-28) = 110.25 + 7
f(-28) = 117.25
So the vertex of the function is (-28, 117.25).
Now, let's move on to the expression (6x^2-5) ( x^7-6) and determine if your expansion is correct.
Using the distributive property, we need to multiply every term of the first parentheses with every term of the second parentheses.
(6x^2-5) ( x^7-6) can be expanded as follows:
6x^2 * x^7 + 6x^2 * (-6) - 5 * x^7 - 5 * (-6)
Simplifying further, we get:
6x^9 - 36x^2 - 5x^7 + 30
So, your expansion is correct: (6x^2-5) ( x^7-6) = 6x^9 - 36x^2 - 5x^7 + 30.
I hope this helps! Let me know if you have any further questions.