question archive Part 1: x-intercepts The x-intercepts occur at point son the function where ##y=0## So, we need to solve ##x^2-4x-12 = 0## The left side factors fairly easily into: ##(x-6)(x+2)=0## So solution occur when ##x-6 = 0 rarr x=6## and ##x+2= 0 rarr x=(-2)## So the x-intercepts are at ##(0,6)## and ##(0,-2)## Part 2: vertex of the parabola The vertex of a simple quadratic parabola occurs when the derivative of the quadratic is equal to ##0##
Part 1: x-intercepts The x-intercepts occur at point son the function where ##y=0## So, we need to solve ##x^2-4x-12 = 0## The left side factors fairly easily into: ##(x-6)(x+2)=0## So solution occur when ##x-6 = 0 rarr x=6## and ##x+2= 0 rarr x=(-2)## So the x-intercepts are at ##(0,6)## and ##(0,-2)## Part 2: vertex of the parabola The vertex of a simple quadratic parabola occurs when the derivative of the quadratic is equal to ##0##
Subject:MathPrice: Bought3
Part 1: x-intercepts The x-intercepts occur at point son the function where ##y=0## So, we need to solve ##x^2-4x-12 = 0## The left side factors fairly easily into: ##(x-6)(x+2)=0## So solution occur when ##x-6 = 0 rarr x=6## and ##x+2= 0 rarr x=(-2)## So the x-intercepts are at ##(0,6)## and ##(0,-2)##
Part 2: vertex of the parabola The vertex of a simple quadratic parabola occurs when the derivative of the quadratic is equal to ##0##. The derivative of the given quadratic is ##(dy)/(dx) = 2x -4## By observation, this is equal to ##0## when ##x=2## When ##x=2## the original equation becomes ##y = (2)^2 -4(2)-12## ##y = -16## Therefore the vertex of this parabola is at ##(2,-16)##
