Given the equation of the curve $y=x^2+x$ and the requirement that the tangent to this curve passes through $(2, -3)$, what are the possible equations of the tangent lines?
I started by differentiating the curve: $ y'=2x+1$. Then I let the point of tangency be $(a, a^2+a)$.
After solving the equations, I ended up finding the equations of the tangents to the curve passing through $(2, -3)$ to be: $y=11x-25$ and $y=-x-1$.
edit: found the points of tangency to be $(5, 30)$ and $(-1, 0)$
Am I correct? Are there more solutions?
$\endgroup$ 13 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$First we have:
$$y = x^2 + x$$ $$y' = 2x + 1$$ $$p = (2,-3)$$ Next we must construct two projective right-angle triangles with the hypotenuse connecting p to points m (on the left) and n (on the right). For m: $$(-x_m + 2)(-y'_m) - 3 = y_m$$ $y'_m$ is negated because we are looking at it from the right and $x_m$ is negated because we want the absolute value of the base. This simplifies to: $$(x_m - 2)y'_m - 3 = y_m$$ Which is the same as the formula for n: $$(x_n - 2)y'_n - 3 = y_n$$ So we can generalize: $$(x - 2)y' - 3 = y$$ But at both m and n the derivative of the curve and the value of y is the same as that for the respective line so, substituting: $$(x - 2)(2x + 1) - 3 = x^2 + x$$ Which, as stated by CaptainObvious is a quadratic with solutions 5 and -1, so the y values are 0 and 30. This allows us to compute the gradients of mp and pn: $$\frac{0 - 3}{2 - -1} = -1$$ $$\frac{30 - -3}{5 - 2} = 11$$ Which are the same values you get if you substitute -1 and 5 into the derivative of the curve. This was the stated condition.
$\endgroup$ $\begingroup$You have an equation for your slope in $y'$ and you have a point. I would advise you to use point-slope form $y-y_0=y'(x-x_0)$.
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