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The differentiate of a function gives the gradient of the original function at the point $x$.

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The tangent to a line is the gradient of the curve at the point, and the normal is the reciprocal of this value (the perpendicular line intersecting the curve).

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The equation of a line can be found in many ways, but given a point and gradient for a straight line, two of the simple manipulations are:

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Using $y=mx+c$ and substituting in to find $c$

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OR

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Using $\\frac{y_2-y_1}{x_2-x_1}=m$ and rearranging.

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Using differentiation to find the tangent and normal to a line at a given point

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Find an equation for the tangent to the curve $\\simplify{y={a[0]}x^2+{b[0]}x+{c[0]}}$ at the point at which it crosses the y-axis.

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$\\frac{dy}{dx}=$ [[0]]

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Therefore, the gradient of the tangent at $x=0$ is [[1]]

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When $x=0$, $y=$ [[2]]

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Hence, the equation for the tangent is:

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So, $y=$ [[3]]

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Find an equation for the normal to the curve $\\simplify{y={a[1]}x^2+{b[1]}x+{c[1]}}$ at the point where $x=\\var{d}$.

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$\\frac{dy}{dx}=$ [[0]]

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Therefore, gradient of the curve at $x=\\var{d}$ is [[1]]

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Meaning the gradient of the normal at $x=\\var{d}$ is [[4]]

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When $x=\\var{d}$, $y=$ [[2]]

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Hence, the equation for the normal is:

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So, $y=$ [[3]]

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Find the following.

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