#### To determine

**To find:** The equation of the tangent lines to the parabola through the points.

#### Answer

The tangent line of curve at (−1,0) is y=−x−1.

The tangent line of curve at (5,30) is y=11x−25.

#### Explanation

**Given:**

The parabola is y=x2+x.

The parabola through the point is (2,−3).

**Derivative rules:**

(1) Power Rule: ddx(xn)=nxn−1

(2) Sum Rule: ddx[f(x)+g(x)]=ddx(f(x))+ddx(g(x))

**Formula used:**

The equation of tangent line at (x1,y1) is, y−y1=m(x−x1) (1)

Here, *m* is the slope of the tangent line at (x1,y1) and m=dydx|x=x1.

**Calculation:**

The derivative of y is dydx, which is obtained as follows.

dydx=ddx(y) =ddx(x2+x)

Apply the sum rule (2).

dydx=ddx(x2)+ddx(x)

Apply the power rule (1) and simplify the expression.

dydx=2x2−1+1x1−1=2x+1

Therefore, the derivative of the parabola y=x2+x is 2x+1.

Given parabola is y=x2+x, that is every point of the parabola is (a,a2+a) of the form.

Obtain the equation of the tangent line to the parabola y=x2+x at the point (a,a2+a).

The slope of the tangent line at (a,a2+a) is,

m=dydx|x=a=2(a)+1=2a+1

Thus, the slope of tangent line at (a,a2+a) is m=2a+1 (2)

Note that, the slope of the line passing through the point (x1,x2) and (y1,y2) is m=y2−y1x2−x1.

The tangent line passes through the point (2,−3) and any point (a,a2+a) to the curve is,

m=a2+a−(−3)a−2=a2+a+3a−2

Therefore, the slope m=a2+a+3a−2 (3)

Compare the equations (2) and (3),

2a+1=a2+a+3a−2

Cross multiply the equation and simplify the terms,

(a−2)(2a+1)=a2+a+32a2+a−4a−2=a2+a+3a2−4a−5=0

Solve the above equation and obtain a=−1 and a=5.

Substitute the values a=−1 and a=5 in (a,a2+a), the points are (−1,0) and (5,30).

Since the slope of the tangent line at (a,a2+a) is m=2a+1, the slope at (−1,0) and (5,30) are −1 and 11, respectively.

Substitute (−1,0) for (x1,y1) and −1 for *m* in equation (1),

y−0=−1(x+1)y=−x−1

The tangent line to the curve at (−1,0) is y=−x−1.

Substitute (5,30) for (x1,y1) and 11 for *m* in equation,

y−30=11(x−5)y−30=11x−55y=11x−55+30y=11x−25

The tangent line of curve at (5,30) is y=11x−25.

**The graph:**

The graph of the curve and tangent line as shown below in Figure 1.

From Figure 1, it is observed that the lines y=11x−25 and y=−x−1 are tangent to the curve y=x2+x and these two lines are intersect at the point (2,−3).

#### To determine

**To show:** There is no line through the point (2,7) that is tangent to the parabola.

#### Explanation

The given parabola is y=x2+x, that is every point of the parabola is (a,a2+a) of the form.

Note that, the slope of the line passing through the point (x1,x2) and (y1,y2) is m=y2−y1x2−x1

The tangent line passing through the point (2,7) and any point to the parabola (a,a2+a),

m=a2+a−(7)a−2=a2+a−7a−2

Therefore, the slope of the tangent line m=a2+a−7a−2.

From part (a), the slope of tangent line at (a,a2+a) is, m=2a+1.

Then, the equation becomes,

2a+1=a2+a−7a−2(2a+1)(a−2)=a2+a−72a2−4a+a−2=a2+a−7a2−4a+5=0

Note that, the quadratic equation Ax2+Bx+C=0 has no real solution if the discriminant B2−4AC<0.

B2−4AC=16−4(1)(5)=16−20=−4

Thus, the equation a2−4a+5=0 has no real solution.

Therefore, it is concluded that there is no line such that the tangent line passing through the point (2,7) to the parabola.

**Graph:**

The graph of the parabola y=x2+x and the point (2,7) is shown below in Figure 2.

From Figure 2, it is observed that there is no tangent line to the parabola through the point (2,7).