How do you find the VERTEX of a parabola $y = \frac{3}{4} x^{2}$ $y= 3/4x^2$ ?

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How do you find the VERTEX of a parabola $y = \frac{3}{4} x^{2}$ $y= 3/4x^2$ ?

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Answer 1 · 2018-04-01T18:11:30

$(0, 0)$ $(0,0)$

Explanation:

In general, the vertex of a parabola in the form $y = a x^{2}$ $y=ax^2$ where $a$ $a$ is any number is $(0, 0),$ $(0,0),$ as the parabola has not been shifted up, down, left, or right at all.

We can prove this by comparing $y = \frac{3}{4} x^{2}$ $y=3/4x^2$ to the standard form of a quadratic, $y = a x^{2} + b x + c$ $y=ax^2+bx+c$

In this case, $a = \frac{3}{4}, b = 0, c = 0$ $a=3/4, b=0, c=0$

The $x -$ $x-$ coordinate of the vertex is given by $- \frac{b}{2 a}$ $-b/(2a)$ . In this case, it would be $(- \frac{0}{2 \cdot \frac{3}{4}}) = 0$ $(-0/(2*3/4))=0$ .

The $y -$ $y-$ coordinate of the vertex is given by plugging in the result of $- \frac{b}{2 a}$ $-b/(2a)$ into the equation. In this case, $- \frac{b}{2 a} = 0, y = \frac{3}{4} (0^{2}) = 0$ $-b/(2a)=0, y=3/4(0^2)=0$

So, the vertex is $(0, 0)$ $(0,0)$

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How do you find the VERTEX of a parabola $y = \frac{3}{4} x^{2}$ $y= 3/4x^2$ ?

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