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.

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