If the parabola y = alpha x^{2} - 6x + beta p | vedclass

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Question

(C) Given the parabola equation: $y = \alpha x^{2} - 6x + \beta \dots (i)$ Since the parabola passes through the point $(0, 2)$,we substitute $x = 0$

Vedclass · Accepted Answer

$\alpha = 2, \beta = 2$

Vedclass · Answer

(C) Given the parabola equation: $y = \alpha x^{2} - 6x + \beta \dots (i)$ Since the parabola passes through the point $(0, 2)$,we substitute $x = 0$ and $y = 2$ into equation $(i)$: $2 = \alpha(0)^{2} - 6(0) + \beta \implies \beta = 2$ Now,differentiate equation $(i)$ with respect to $x$: $\frac{dy}{dx} = 2\alpha x - 6$ The tangent at $x = \frac{3}{2}$ is parallel to the $X$-axis,which means the slope $\frac{dy}{dx} = 0$ at $x = \frac{3}{2}$: $\left(\frac{dy}{dx}\right)_{x = 3/2} = 2\alpha \left(\frac{3}{2}\right) - 6 = 0$ $3\alpha - 6 = 0 \implies 3\alpha = 6 \implies \alpha = 2$ Thus,the values are $\alpha = 2$ and $\beta = 2$.

If the parabola $y = \alpha x^{2} - 6x + \beta$ passes through the point $(0, 2)$ and has its tangent at $x = \frac{3}{2}$ parallel to the $X$-axis,then:

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