Determine the quadratic curve `y=f(x)` if it touches the line `y=x` at the point `x=1` and passes through the point `(-1,\ 0)` .

Answer

The required curve is: `y=frac{x^{2}}{4}+frac{x}{2}+frac{1}{4}`

Explanation

\begin{aligned}
&\text { Let the required quadratic curve be } \\
&\text { It passes through }(-1,0) \text {. Therefore, } \\
&0=a-b+c \ldots(\text { ii) } \\
&\text { Since the line } y=x \text { touches (i) at } x=1 \\
& \therefore (\text { slope of the tangent at } x=1)=(\text { slope of the line } y=x) \\
&\left(\frac{d y}{d x}\right)_{x-1}=1 \\
& 2 a+b=1 \ldots \text { (iii) }\left[\because y=a x^{2}+b x+c \therefore \frac{d y}{d x}=2 a x+b\right] \\
...