Answer:
Step-by-step explanation:
We can answer this in two ways: Differentiation and Graphing.
<u>Differentitate:</u>
The first derivative of a function yields a function that provides the slope for any point on the line of the original function. The slope of a vertex is zero, so we can set the first derivative to 0 and solve for x.
f(x) = 3(x-1)^2 + 4
f(x) = 3(x-1)(x-1) + 4
f(x) = 3(x^2 - 2x + 1) + 4
f(x) = 3x^2 - 6x + 7
f'(x) = 6x -6
Set this = 0 and find x:
0 = 6x -6
x = 1
The value of y when x=1 in the original equation is:
f(1) = 3(1-1)^2 + 4
y = 4
The vertex is (1,4)
<u>Graph:</u>
You can use DESMOS to plot the function. The result is attached. Look for the vertex and read the coordinates. (1,4) seems to work.
Answer: C
Step-by-step explanation:
They both end up equaling 4x+16 so they cancel each other out
I'm not sure how to give you solid proof, like an equation or something, but you can see that out of the four coordinates, two of them have the y-value of 4 and two of them have the y-value of 2. Also, there is 1.5 in between each x-value that is paired with the same y-value... if that makes sense. So for (0.5, 2) and (3, 2), there is 1.5 in between 0.5 and 3. I hope this helps! Sorry it's a bit of a weird answer
Answer:
(xy - 4)(xy + 4).
Step-by-step explanation:
The is the difference of 2 squares.
a^2 - b^2 = (a - b)(a + b: so we have:
x^2y^2 - 16 = (xy - 4)(xy + 4)