What is the vertex of the quadratic function y = -7x^2 + 28x - 5?

Question

What is the vertex of the quadratic function y = -7x^2 + 28x - 5? 

 

Thank you! 

Answer 1

It is a process called completing the square.

y = -7x^2 + 28x - 5

first simplify the x^2 term just to make the math easier
-(1/7)y = x^2 -4x + 5/7

and just to make the process clearer, move the 5/7 to the other side of the equals sign. (Not necessary in practice, just good procedure to demonstrate the process.)
-(1/7)y - 5/7 = x^2 -4x

if the right side of the equation were a natural square (a + b)^2, then it would have the form a^2 + 2ab + b^2.

in this instance a = x and 2ab = -4x

so b = -2, and therefore b^2 =(-2)^2 = 4

adding 4 to the right side of the equation makes it a perfect square, just remember to add it to the left side as well.
-(1/7)y - 5/7 + 4 = x^2 -4x + 4 = (x - 2)^2  Solve for y
-(1/7)y = (x-2)^2 -(23/7)
y = -7 (x-2)^2 + 23

the axis of symmetry is x=2 because that turns the x^2 coefficient to 0.  Finally, the graph has been translated up 28 along the y-axis.

thus the vertex is (2,23)