Tangents and normals

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A quadratic function y=f(x) if it touches the line y=x at the point x=1 and passes through the point (-1, 0)

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Solution

$\begin{array}{l} y = f (x) = a x^{2} + b x + c \\ f^{'} (x) = 2 a x + b = 1 \\ at x = 1 \\ \Rightarrow 2 a + b = 1 - - - - (1) \end{array}$
$\begin{array}{l} x = 1, y = 1 \\ \Rightarrow f (1) = 1 \\ \Rightarrow a + b + c = 1 - - - - (2) \\ Solving (1) & (2) \end{array}$
$\begin{array}{l} curve passes through (- 1, 0) \\ a = \frac{1}{4} \\ b = \frac{1}{2} \\ c = \frac{1}{4} \\ F (= \frac{x^{2}}{4} + \frac{x}{2} + \frac{1}{4}) \\ = \frac{x^{2} + 2 x + 1}{4} = \frac{1}{4} (x + 1)^{2} \end{array}$

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Tangents and normals

Remember concepts with our Masterclasses.

A quadratic function y=f(x) if it touches the line y=x at the point x=1 and passes through the point (-1, 0)

y=f(x)=ax2+bx+cf′(x)=2ax+b=1 at x=1⇒2a+b=1----1x=1,y=1⇒f(1)=1⇒a+b+c=1----2 Solving (1)&(2) curve passes through (−1,0)a=14b=12c=14F=x24+x2+14=x2+2x+14=14(x+1)2

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