How do you graph and label the vertex and axis of symmetry $y = \frac{1}{3} x^{2} + 3 x - 2$ $y=1/3x^2+3x-2$ ?

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How do you graph and label the vertex and axis of symmetry $y = \frac{1}{3} x^{2} + 3 x - 2$ $y=1/3x^2+3x-2$ ?

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Answer 1 · 2018-08-04T05:05:00

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Explanation:

Given -

$y = \frac{1}{3} x^{2} + 3 x - 2$ $y=1/3x^2+3x-2$

$x = \frac{- b}{2 a} = \frac{- 3}{2 \times \frac{1}{3}} = \frac{- 3}{\frac{2}{3}} = \frac{- 3 \times 3}{2} = \frac{- 9}{2}$ $x=(-b)/(2a)=(-3)/(2 xx1/3)=(-3)/(2/3)=(-3 xx 3)/2=(-9)/2$

$A t x = \frac{- 9}{2}; y = \frac{1}{3} {(\frac{- 9}{2})}^{2} + 3 (\frac{- 9}{2}) - 2$ $At x=(-9)/2; y=1/3((-9)/2)^2+3((-9)/2)-2$

$y = \frac{1}{3} (\frac{81}{4}) - \frac{27}{2} - 2 = \frac{81}{12} - \frac{27}{2} - 2 = \frac{81 - 162 - 24}{12} = \frac{- 105}{12} = \frac{- 35}{4}$ $y=1/3(81/4)-27/2-2=81/12-27/2-2=(81-162-24)/12=(-105)/12=(-35)/4$

Vertex $(- \frac{9}{2}, - \frac{35}{4})$ $(-9/2,-35/4)$

Axis of sysmetry $x = - \frac{9}{2}$ $x=-9/2$

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How do you graph and label the vertex and axis of symmetry $y = \frac{1}{3} x^{2} + 3 x - 2$ $y=1/3x^2+3x-2$ ?

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How do you graph and label the vertex and axis of symmetry $y = \frac{1}{3} x^{2} + 3 x - 2$ $y=1/3x^2+3x-2$ ?