What is the the vertex of #y = 2(x-4)^2+3x-12#?

Question

Savannah Anderson · Accepted Answer

( #13/4#, #-9/8# )

First off let's simplify the whole equation and collect like terms. After squaring (x-4) and multiplying the result by 2 we must add 3 to the x term and subtract 12 from the constant. Collecting everything gives us: #f(x)# = #2 x^2 - 13 x + 20# The fastest way to find the vertex of a parabola is to find the point where it's derivative equals 0. This is because the slope of the tangent line is equal to 0 anytime the graph of a parabola forms a horizontal line. If you have not done calculus do not worry about this and simply KNOW that the derivative when = 0 will give you the x value of the vertex. The derivative of f(x) = #f'(x)# where #f'(x)# = #4x-13# #f'(x)# = 0 at the point #(13/4) # Plug #(13/4)# back into #f(x)# to get #f(13/4)# which gives #-9/8#. Therefore the answer is found to be: x = #13/4# and y = #-9/8# therefore: Vertex = (#13/4#,#-9/8#) Note: I understand some of you may not have done derivatives yet. My honest answer is to youtube derivatives of quadratic equations as this method will save you tons of time, and understanding derivatives of quadratic or linear equations is very straightforward using the power rule.

Avery Alexander · Answer

undefined The vertex of the quadratic function $ y = 2(x-4)^2+3x-12 $ is (4, -12).