Suppose the tangent to the parabola y=x^2+px+ | vedclass

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(C) The equation of the parabola is given by $y = x^2 + px + q$.Since the parabola passes through the point $(0, 3)$,we substitute $x = 0$ and $y = 3$

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(C) The equation of the parabola is given by $y = x^2 + px + q$.Since the parabola passes through the point $(0, 3)$,we substitute $x = 0$ and $y = 3$ into the equation:$3 = (0)^2 + p(0) + q \implies q = 3$.Next,we find the derivative of $y$ with respect to $x$ to determine the slope of the tangent:$\frac{dy}{dx} = 2x + p$.The slope of the tangent at $(0, 3)$ is given by evaluating the derivative at $x = 0$:$\left(\frac{dy}{dx}\right)_{x=0} = 2(0) + p = p$.Given that the slope of the tangent is $-1$,we have $p = -1$.Finally,we calculate $p + q$:$p + q = -1 + 3 = 2$.

Suppose the tangent to the parabola $y=x^2+px+q$ at $(0,3)$ has slope $-1$. Then,$p+q$ equals

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