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)dot`

Answer

the equation is `y = 0.25x^2 + 0.5x + 0.25` or `(x + 1)^2/4`.

Explanation

As `f(x)` is quadratic, we can write `y = ax^2 + bx + c`

The gradient of the tangent at any point on the graph corresponding to x is `f’(x)`[ its derivative `dy/dx = 2ax + b`]

Hence,` 2a + b = 1`, which is the gradient of `y = x`.

Also as the point `(1,1)` lies on the curve, we have `a + b + c = 1`.

And, as `(-1,0)` lies on the curve, `a - b + c = 0`.

Now solving the equations

`2a + b = 1`, `a + b + c = 1` and `a - b + c = 0` to find `a = 0.25, b = 0.5` and `c = 0.25`.

Hence, the equation is `y = 0.25x^2 + 0.5x + 0.25` or `(x + 1)^2/4`.