What is the vertex of $y = 3 x^{2} + 2 x + 5$ $y=3x^2+2x+5$ ?

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Answer 1 · 2017-10-04T16:31:56

Vertex $\to (x, y) = (- \frac{1}{3}, \frac{14}{3})$ $->(x,y)=(-1/3,14/3)$

Given: $y = 3 x^{2} + 2 x + 5$ $y=3x^2+2x+5$

This is part of the process of completing the square.

Write as $y = 3 (x^{2} + \frac{2}{3} x) + 5$ $y=3(x^2color(red)(+2/3)x)+5$

To complete the square you would 'do other things' to this. I am not going to do that!

$x_{vertex} = (- \frac{1}{2}) \times (+ \frac{2}{3}) = - \frac{1}{3}$ $x_("vertex")=(-1/2)xx( color(red)(+2/3))= -1/3$

Substitute for $x$ $x$ to determine $y_{vertex}$ $y_("vertex")$

$y_{vertex} = 3 {(- \frac{1}{3})}^{2} + 2 (- \frac{1}{3}) + 5$ $y_("vertex")= 3(-1/3)^2+2(-1/3)+5$

$y_{vertex} = + \frac{1}{3} - \frac{2}{3} + 5 = 4 \frac{2}{3} \to \frac{14}{3}$ $y_("vertex")= +1/3-2/3+5 = 4 2/3 -> 14/3$

Vertex $\to (x, y) = (- \frac{1}{3}, \frac{14}{3})$ $->(x,y)=(-1/3,14/3)$

What is the vertex of y=3x2+2x+5y=3x^2+2x+5?