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Sagot :
Step 1: Write out the function
[tex]f(x)=-x^2+6x-7^{}[/tex]Step 2: Differentiate the function in step 1 with respect to x
[tex]f^{\prime}(x)=-2x+6[/tex]Step 3: Find the gradient, n, of the tangent line through Q.
the gradient of the tangent line throught Q is given by the value of f'(x) at the point (4,1)
[tex]\begin{gathered} \text{ In this case,} \\ x=4 \\ \text{ Therefore} \\ n=f^{\prime}(4)=-2(4)+6=-8+6=-2 \end{gathered}[/tex]Step 4 (a): Find the gradient of the line L.
Since line L is perpendicular to the tangent line of Q, then the gradient, m, of the line L is given by
[tex]\begin{gathered} m=-\frac{1}{n} \\ \text{ Thus} \\ m=-\frac{1}{-2}=\frac{1}{2} \end{gathered}[/tex]Step 5: Write out the formula for finding the equation of a line through the point (x1,y1)
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \text{ wh}ere \\ m\text{ is the gradient of the line} \end{gathered}[/tex]Step 6 (b): Find the equation of line L
[tex]\begin{gathered} y-1=\frac{1}{2}(x-4) \\ y=\frac{x}{2}-1 \end{gathered}[/tex]Hence the equation of line L is: y = 1/2 x - 1
Step 7 (c): Find the coordinates of R
At R, line L intersects the curve.
Therefore,
[tex]\begin{gathered} -x^2+6x-7=\frac{1}{2}x-1 \\ The\text{ refore} \\ x^2-\frac{11}{2}x+6=0 \\ (x-\frac{11}{4})^2-(-\frac{11}{4})^2+6=0 \\ (x-\frac{11}{4})^2=\frac{25}{16} \\ x-\frac{11}{4}=\pm\frac{5}{4} \\ x=\frac{11+5}{4},\frac{11-5}{4}=4,\frac{3}{2} \end{gathered}[/tex]at x = 3/2
[tex]f(\frac{3}{2})=4.25[/tex]Hence, the coordinates of R is (3/2, 4 1/4)
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