Ineed with my khan's can someone me

answer: (-26,10)

explanation

a normal line to a point on a curve is a line that is perpendicular to the tangent line to the same point on the curve. we can use differential calculus to find the derivative of as that will get us the tangent line slopes of the curve.

for notation's sake, let be the function such that

we can first calculate the points on the curve at x = 1 and x = 2:

so points (1,1) and (2,6) are on this curve.

now we can find the derivative to get the tangent line slopes. differentiating with the power and sum rules for derivatives, we get

finding the slope of the tangent line to the curve at x = 1 and x = 2:

since the normal line is perpendicular to the tangent line, and perpendicular lines are negative reciprocals of each other, the slope of the normal

using the point-slope form for a line, , the normal to the curve at the point (1,1) has the equation

and the normal to the curve at the point (2,6) has the equation

to find where these normals intersect, equate their equations to each other and solve for x to find the x-coordinate of intersection:

to get rid of fractions, multiply both sides of the equation by -3 and then by 7 so that the denominators cancel out, then follow through with simplifying.

to find the y-coordinate, use either normal equation

(-26,10) is the coordinates of the point where these normals intersect.