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

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

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Answer 1 · 2016-01-12T10:55:11

The Vertex is at ( $\frac{1}{3}, - 4 \frac{2}{3}$ $1/3, -4 2/3$ )

Explanation:

This is the equation of the Parabola opens down as co-efficient of $x^{2}$ $x^2$ is negative. Comparing with the General equation ( $a x^{2} + b x + c$ $ax^2+bx+c$ )
we get a = (-3) ; b= 2 ; c =(-5) Now we know x-co-ordinate of the vertex is equal to -b/2a. so $x_{1} = - \frac{2}{2 \cdot (- 3)}$ $x_1=-2/(2*(-3))$ or $x_{1} = \frac{1}{3}$ $x_1 = 1/3$
Now putting the value of x =1/3 in the equation we get $y_{1} = - 3 . {(\frac{1}{3})}^{2} + 2 \cdot (\frac{1}{3}) - 5$ $y_1=-3.(1/3)^2 +2*(1/3) -5$ or $y_{1} = - \frac{14}{3}$ $y_1 = -14/3$ or $y_{1} = - (4 \frac{2}{3})$ $y_1=-(4 2/3)$ So The Vertex is at $(\frac{1}{3}, - 4 \frac{2}{3})$ $(1/3, -4 2/3)$

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

1 Answer

Explanation:

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

1 Answer

Explanation:

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