Answer
(a) The equations of the two tangent lines are:
$y = -x-1$
$y = 11x-25$
(b) There is no line through the point $(2, 7)$ that is tangent to the parabola.
The sketch shows that the point $(2,7)$ is above the parabola.
Work Step by Step
(a) $y = x^2+x$
$\frac{dy}{dx} = 2x+1$
At a point where a tangent line touches a graph, the slope of the graph at that point is equal to the slope of the tangent line.
We can find the two points where straight lines through the point $(2,-3)$are tangent to the graph:
$y-(-3) = m(x-2)$
$y+3 = m(x-2)$
$x^2+x+3 = (2x+1)(x-2)$
$x^2+x+3 = 2x^2-3x-2$
$x^2-4x-5=0$
$(x-5)(x+1)=0$
$x = -1, 5$
When $x = -1~~$ then $~~m = \frac{dy}{dx} = 2(-1)+1 = -1$
We can find the equation of this tangent line:
$y-(-3) = m(x-2)$
$y+3 = (-1)(x-2)$
$y = -x-1$
When $x = 5~~$ then $~~m = \frac{dy}{dx} = 2(5)+1 = 11$
We can find the equation of this tangent line:
$y-(-3) = m(x-2)$
$y+3 = (11)(x-2)$
$y = 11x-25$
(b) We can try to find a point where a straight line through the point $(2,7)$ is tangent to the graph:
$y-7 = m(x-2)$
$x^2+x-7 = (2x+1)(x-2)$
$x^2+x-7 = 2x^2-3x-2$
$x^2-4x+5=0$
Using the quadratic formula:
$x = \frac{4\pm \sqrt{(-4)^2-4(1)(5)}}{2(1)}$
$x = \frac{4\pm \sqrt{16-20}}{2}$
Since there is a negative number inside the square root, there are no solutions.
On the graph, we can see that the point $(2,7)$ is above the parabola. Since the parabola is concave up, there are no tangent lines that go through the point $(2,7)$
