Home > Algebra > Quadratic Functions and Their Graphs

What is the vertex of # y= (x-3)^2-2x^2-4x-9#?

Answer 1

Hazel Anglin

#-5,25)#

#"first express in standard form "y=ax^2+bx+c color(white)(x);a!=0#

#"expand "(x-3)^2" using Foil and collect like terms"#

#y=x^2-6x+9-2x^2-4x-9#

#color(white)(y)=-x^2-10x#

#"the x-coordinate of the vertex is on the axis of"# #"symmetry passing through the midpoint of the zeros"#

#"let y = 0"#

#rArr-x^2-10x=0#

#rArr-x(x+10)=0#

#rArrx=0,x=-10larrcolor(red)" are the zeros"#

#x_(color(red)"vertex")=(0-10)/2=-5#

#y_(color(red)"vertex")=-(-5)^2-10(-5)=25#

#rArrcolor(magenta)"vertex "=(-5,25)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Eva Abramson

To find the vertex of the quadratic function (y = (x - 3)^2 - 2x^2 - 4x - 9), you need to rewrite it in the standard form (y = ax^2 + bx + c), where (a), (b), and (c) are constants. Then, you can use the formula for the x-coordinate of the vertex, given by (x = \frac{-b}{2a}), and substitute it back into the function to find the y-coordinate.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.