How do you find the vertex and intercepts for #y = 3(x - 2)^2 + 5#?

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Answer 1 · 2017-12-30T03:53:11

Vertex: #(2,5)#
y-intercept: #(0,17)#
x-intercept is undefined in ℝ

First, expand #y=3(x-2)^2+5# into #y=3x^2-12x+17#

The x-vertex of a quadratic expression (#y=ax^2+bx+c#) is given by #-b/(2a)#.

So, the x-vertex for #y=3x^2-12x+17# will be #(-(-12))/(2*3)=12/6=2#

Plugging in the x-vertex into our original expression gives us the y-vertex, which is #3*4-24+17=5#.

So, our vertex is at #(2,5)#.

The y-intercept is given when #x=0# in the expression, so it is #17#.

y-intercept: #(0,17)#

graph{y=3(x-2)^2+5 [-11.26, 11.545, 0.32, 11.72]}

From this graph of #y=3x^2-12x+17#, it never hits the x-axis, so there is no x-intercept.