If the curve y=ax^2-6x+b passes through (0,4) | vedclass

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(A) Given the curve equation is $y = ax^2 - 6x + b$.Since the curve passes through $(0, 4)$,we substitute $x=0$ and $y=4$ into the equation:$4 = a(0)^

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$2, 4$

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(A) Given the curve equation is $y = ax^2 - 6x + b$.Since the curve passes through $(0, 4)$,we substitute $x=0$ and $y=4$ into the equation:$4 = a(0)^2 - 6(0) + b \implies b = 4$.Now,the derivative of the curve is $\frac{dy}{dx} = 2ax - 6$.The tangent is parallel to the $x$-axis at $x = \frac{3}{2}$,which means $\frac{dy}{dx} = 0$ at $x = \frac{3}{2}$.$2a(\frac{3}{2}) - 6 = 0 \implies 3a - 6 = 0 \implies 3a = 6 \implies a = 2$.Thus,the values are $a = 2$ and $b = 4$.

If the curve $y=ax^2-6x+b$ passes through $(0,4)$ and has its tangent parallel to the $x$-axis at $x=\frac{3}{2}$,then the values of $a$ and $b$ respectively are

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