Question

at what point does the line normal to the graph of y = 1 + x+2x^2 at (1,4) intersect the parabola a second time? hint: the normal line is perpendicular to the tangent line. if two lines are perpendicular their slopes are negative reciprocals -- i.e. if the slope of the first line is m then the slope of the second line is -1/m

Explanation:

Step1: Find the derivative of the function

The function is $y = 1+x + 2x^{2}$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $y^\prime=\frac{d}{dx}(1)+\frac{d}{dx}(x)+\frac{d}{dx}(2x^{2})=0 + 1+4x$.

Step2: Find the slope of the tangent at the point $(1,4)$

Substitute $x = 1$ into $y^\prime$. So $y^\prime(1)=1 + 4\times1=5$.

Step3: Find the slope of the normal line

Since the normal line is perpendicular to the tangent line, if the slope of the tangent line is $m = 5$, the slope of the normal line $m_{n}=-\frac{1}{5}$.

Step4: Write the equation of the normal line

Using the point - slope form of a line $y - y_{1}=m_{n}(x - x_{1})$, with $(x_{1},y_{1})=(1,4)$ and $m_{n}=-\frac{1}{5}$, we get $y - 4=-\frac{1}{5}(x - 1)$, which simplifies to $y=-\frac{1}{5}x+\frac{1}{5}+4=-\frac{1}{5}x+\frac{21}{5}$.

Step5: Find the intersection point with the parabola

Set $1+x + 2x^{2}=-\frac{1}{5}x+\frac{21}{5}$. Multiply through by 5 to clear the fraction: $5 + 5x+10x^{2}=-x + 21$. Rearrange to get a quadratic equation: $10x^{2}+6x - 16 = 0$. Divide by 2: $5x^{2}+3x - 8 = 0$.

Step6: Solve the quadratic equation

Factor the quadratic equation $5x^{2}+3x - 8=(5x + 8)(x - 1)=0$. We know $x = 1$ is the given point. Solving $5x+8 = 0$ gives $x=-\frac{8}{5}$.

Step7: Find the corresponding y - value

Substitute $x = -\frac{8}{5}$ into $y = 1+x + 2x^{2}$. $y=1-\frac{8}{5}+2\times(\frac{64}{25})=\frac{25 - 40+128}{25}=\frac{113}{25}$.

Answer:

$(-\frac{8}{5},\frac{113}{25})$