A calculus problem by Parth Lohomi

We note that the line (red) tangent to the parabola cuts the $x$ -axis and $y$ -axis at $(x_1, y_1) = \left(\frac{1}{2}, 0\right)$ and $(x_2, y_2) = (0,-1)$ respectively. The gradient $m$ of the line is given by:

$\begin{aligned} m & = \frac{y_2-y_1}{x_2-x_1} = \frac{-1-0}{0-\frac12} = \boxed{2} \end{aligned}$

It can be seen that the parabola is $y = x^2$ . Therefore its gradient at $x$ is given by $\dfrac {dy}{dx} = 2x$ . Since the tangent touches the parabola at $x=1$ , therefore the gradient is $\boxed{2}$ .

We can see that the blue graph represents $f(x)=x^2$ . The slope of the tangent line is $f'(x_0)$ where $x_0$ is the point where the line is tangent. We can see that $x_0=1$ so $f'(x)=2*x => f'(1)=2$ so the correct answer is 2.

we can use the equation y=mx+c to find the slope here the red line cuts y- axis at (0,-1) therefore,c = -1 & the line touches the point (1,1) so x=1 & y=1 y=mx+c (1)=m(1)+(-1) 1=m -1 hence, m=2

We have: $\begin{cases}y=mx-1\\y=x^2\\ \end{cases}$ so $mx-1=x^2 \newline x^2 - mx + 1 = 0 \newline$ Here x has only one solution so $\Delta = 0 \newline b^2-4ac=0 \newline m^2-4=0 \newline m^2 = 4 \newline$ m must be greater than 0 so $m = \boxed{2}.$