The coordinates of point R are [tex](x,y) = \left(-\frac{31}{5} ,\frac{33}{5} \right)[/tex].
How to determine the point of intersection of a system of equations
In this exercise we must determine the equation of the normal and tangent lines before finding out the point of intersection. By calculus we know that slope of the line tangent to a point of the curve ([tex]m[/tex]) is equal to the first derivative of the function evaluated at that point.
[tex]m = 5\cdot x^{4}-24\cdot x^{2}+16[/tex] (1)
In addition, the slope of the line normal to the curve ([tex]m_{\perp}[/tex]) is determined by following geometric expression:
[tex]m_{\perp} = -\frac{1}{m}[/tex] (2)
Then, we proceed to determine the linear equations:
Point P ([tex]P(x,y) = (1,9)[/tex])
[tex]m = 5\cdot (1)^{4}-24\cdot (1)^{2}+16[/tex]
[tex]m = -3[/tex]
[tex]m_{\perp} = \frac{1}{3}[/tex]
And the intercept of the function is:
[tex]b = y-m\cdot x[/tex]
[tex]b = 9-\left(\frac{1}{3} \right)\cdot (1)[/tex]
[tex]b = \frac{26}{3}[/tex]
Point Q ([tex]Q(x,y) = (-1,-9)[/tex])
[tex]m = 5\cdot (-1)^{4}-24\cdot (-1)^{2}+16[/tex]
[tex]m = -3[/tex]
And the intercept of the function is:
[tex]b = y-m\cdot x[/tex]
[tex]b = -9-\left(-3 \right)\cdot (-1)[/tex]
[tex]b = -12[/tex]
The system of equations is represented by following two expressions:
[tex]y = \frac{1}{3}\cdot x +\frac{26}{3}[/tex] (3)
[tex]y=-3\cdot x -12[/tex] (4)
Then, the coordinates of point R are [tex](x,y) = \left(-\frac{31}{5} ,\frac{33}{5} \right)[/tex]. [tex]\blacksquare[/tex]
To learn more on systems of equations, we kindly invite to check this verified question: https://brainly.com/question/20379472
