What is the function f(x)=3(x^2-8x)+10 written in vertex form

y = 3(x - 4)² - 38

Step-by-step explanation:

We need to complete the square on

   y = 3(x² - 8x) + 10

A quadratic equation is in the form of y = ax² + bx + c.  To complete the square, take half of the b term (here, the b term is -8), then square it...

-8/2 = -4

(-4)² = 16

Now add and subtract that from the equation...

y = 3(x² - 8x + 16 - 16) + 10

Now pull out the -16 from the parenthesis, be careful though, there is a  multiplier of 3 in front of the parenthesis, so it come out as a positive -48

y = 3(x² - 8x + 16) + 10 - 48

x² - 8x + 16 is a perfect square trinomial (we did this by completing the square), so it factors to (x - 4)², and 10 - 48 = -38, so our equation becomes...

y = 3(x - 4)² - 38

This is now in vertex form, which is either the minimum or maximum.

Vertex form is

y = a(x - h)² + k, where (h, k) is the vertex.  If a > 0, then the vertex is a minimum, if a < 0, then the vertex is a maximum.

Our vertex is (4, -38)

pls ask

the terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. a recursive formula for a geometric sequence with common ratio r is given by an=ran−1 a n = r a n − 1 for n≥2 n ≥ 2 .

step-by-step explanation: