How do you find the intercept and vertex of #y = x^2 - 4x - 2#?

Question

Savannah Anderson · Accepted Answer

#"vertex "=(2,-6)," intercepts "=2+-sqrt6# #"The equation of a parabola in "color(blue)"vertex form"# is. #•color(white)(x)y=a(x-h)^2+k# #"where "(h,k)" are the coordinates of the vertex and a"# #"is a multiplier"# #"to obtain this form "color(blue)"complete the square"# #y=x^2+2(-2)x+4-4-2# #y=(x-2)^2-6larrcolor(red)"in vertex form"# #color(magenta)"vertex "=(2,-6)# #"to obtain the x-intercepts let y = 0"# #(x-2)^2-6=0# #(x-2)^2=6# #color(blue)"take the square root of both sides"# #x-2=+-sqrt6larrcolor(blue)"note plus or minus"# #"add 2 to both sides"# #x=2+-sqrt6larrcolor(red)"exact values"#

Genesis Allen · Answer

undefined To find the intercepts of the quadratic function $ y = x^2 - 4x - 2 $, you can set $ y $ equal to zero and solve for $ x $. These solutions will give you the $ x $-coordinates of the intercepts.

1. **$ x $-intercepts (or roots)**:
   Set $ y = 0 $ and solve for $ x $ using the quadratic formula or factoring.

2. **$ y $-intercept**:
   Substitute $ x = 0 $ into the equation to find the $ y $-coordinate of the intercept.

To find the vertex of the parabola represented by the function, you can use the formula for the $ x $-coordinate of the vertex, which is given by $ x = -\frac{b}{2a} $, where $ a $ and $ b $ are the coefficients of the quadratic equation.

Once you find the $ x $-coordinate of the vertex, substitute it back into the equation to find the corresponding $ y $-coordinate. So, the vertex will be $(x, y)$.