How do you find the vertex and intercepts for #y = x^2 - 4x + 4#?

Question

Avery Alexander · Accepted Answer

Vertex: #(2,0)#
y-intercept: #4#
x-intercept: #2# The general vertex form for a parabola is: #color(white)("XXX")y=(x-color(red)(a))^2+color(blue)(b)# for a parabola with vertex at #(color(red)(a),color(blue)(b))# #y=x^2-4x+4# #color(white)("X")=(x-color(red)(2))^2+color(blue)(0)# is therefore a parabola with vertex at #(color(red)(2),color(blue)(0))# #bar(color(white)("XXXXXXXXXXXXXXXXXXXXXXXXXXXX"))# The y intercept is the value of #y# when #x=0 Given #y=x^2-4x+4# when #x=0# #color(white)("XXX")y=4# #bar(color(white)("XXXXXXXXXXXXXXXXXXXXXXXXXXXX"))# The x intercept is the value(s) of #x# when #y=0# Given #y=x^2-4x-4# when #y=0# #color(white)("XXX")0=x^2-4x+4# #color(white)("XXX")x^2-4x+4=0# #color(white)("XXX")(x-2)^2=0# #color(white)("XXX")(x-2)=0# #color(white)("XXX")x=2#

Jameson Allen · Answer

undefined To find the vertex of the parabola represented by the equation $y = x^2 - 4x + 4$, use the formula $x = -\frac{b}{2a}$, where $a$ is the coefficient of $x^2$ (which is 1) and $b$ is the coefficient of $x$ (which is -4). The x-coordinate of the vertex is $x = -\frac{-4}{2(1)} = 2$. To find the y-coordinate, substitute the x-coordinate into the equation: $y = (2)^2 - 4(2) + 4 = 4 - 8 + 4 = 0$. So, the vertex is (2, 0). To find the x-intercepts, set $y = 0$ and solve for $x$: $0 = x^2 - 4x + 4$. This equation factors to $(x - 2)^2 = 0$, so the only solution is $x = 2$. Therefore, the parabola intersects the x-axis at $x = 2$. Since the vertex is also at $x = 2$, the x-intercept and the vertex coincide.