How do you graph and label the vertex and axis of symmetry $y = \frac{1}{2} x^{2} + 4 x + 10$ $y=1/2x^2+4x+10$ ?

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Answer 1 · 2017-11-29T13:14:41

Refer to the graph available with this answer.

We have the quadratic equation

$y = (\frac{1}{2}) x^{2} + 4 x + 10$ $y = (1/2)x^2 + 4x +10$

Axis of Symmetry is given by $\frac{- b}{2 a}$ $(-b)/(2a)$

So, for our quadratic, Axis of Symmetry = $\frac{- 4}{(2 \cdot (\frac{1}{2}))}$ $(-4)/((2*(1/2))$ ,
which gives us $(- 4)$ $(-4)$

Substitute this value of Axis of Symmetry in our original equation to obtain the Vertex.

$y = (\frac{1}{2}) {(- 4)}^{2} + 4 (- 4) + 10$ $y = (1/2)(-4)^2 + 4(-4) + 10$

$y = (\frac{1}{2}) 16 - 16 + 10$ $y = (1/2)16 -16 + 10$

$y = 2$ $y = 2$

So, our Vertex is at $(- 4, 2)$ $(-4, 2)$

Refer to the labeled graph attached with Vertex and Axis of Symmetry identified.
enter image source here

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