Short Answer: Yes, you are correct.
Long Answer:
In this case, we know we need to find the lines tangent to the curve that go through some point, so I would begin by using the elementary point-slope equation to figure out the equation for y in the original function that meets those conditions.
$$y - y_1 = m(x - x_1)$$
The derivative is the slope at some x on the curve, so plug its function in for m. Use the points given in the problem for $x_1$ and $y_1$ respectively:
$$y - (-3) = (2x + 1)(x - 2)$$
Simplify:
$$y + 3 = 2x^2 - 3x - 2$$
Solve for y:
$$y = 2x^2 - 3x - 5$$
Now that we have what y must be in relation to x for the tangent line to meet the conditions given, we can plug it in to the original equation to find the desired values of x on the curve:
$$2x^2 - 3x - 5 = x^2 + x$$
Subtract $x^2 + x$ from both sides:
$$x^2 - 4x - 5 = 0$$
From the quadratic formula, we know that:
$$x = \frac{4 ± \sqrt{36}}{2}$$
Therefore $x = 5$ and $x = -1$, which when plugged into your original function, gives the points you calculated in your answer.
let the point where tangent is drawn to curve be $(a,b)$
so slope of tangent at that point is $\mathrm{\dfrac{dy}{dx}}=2x=2a$
so the equation of tangent at point $(a,b)$:
$$y-b=2a(x-a)\,\,\,\,\,\,\,\,(3)$$ To find to find x-intercept we put $y=0$ (why? you can ask in comments)$$
$$0-b=2a(x-a)$$$$\implies x=\frac{-b}{2a}+a=2\,\,\,\,\,\,\,\,\ (1)$$(because x- intercept i.e $x=2$)
$$\implies b=2a^2-4a$$
$(a,b)$ lie on the parabola so they must satisfy the parabola equation $y=x^2$
so,$$b=a^2\,\,\,\,\,\,\,\,\,(2)$$
on solving equation $1$ and $2$, we get two values of $a$ as $0,4$
now you know $a$, you can find $b$ then plug values in the equation $(3)$. And you will get two equations and that solves the problem :D
Best Answer
The equation of the tangent line is absolutely correct.
To get an idea of how lines transform on changing $m$ and $c$ in the equation $y=mx+c$, I would recommend playing with this
It will be clear after this demonstration that parallel lines have the same slope.
So, let the equation of your new line be $y=x+c$. To find $c$ , Satisfy the point $(2,-3)$ in the line. You have the answer! Cheers :)