Question

Consider the parabola y = 7x − x2. (a) find the slope of the tangent line to the parabola at the point (1, 6).

Answer 1

y = 5x + 1

To obtain the equation we require it's slope (m) and a point on it (a , b)

The slope m = $\frac{dy}{dx}$ at x = 1

$\frac{dy}{dx}$ = 7 - 2x

at x = 1 : $\frac{dy}{dx}$ = 7 - 2 = 5

the equation in the form y - b = m(x - a) where m = 5 and (a,b) = (1,6)

y - 6 = 5(x-1)

y - 6 = 5x - 5

y = 5x + 1 is the equation of the tangent.

Answer 2

The slope of the tangent line to the parabola at the point (1, 6) is 5 while the equation is y = 5x + 1

The formula for the equation of a line in point-slope form is expressed as;

y - y0 = m(x-x0) where:

• m is the slope
• (x0, y0) is the point on the line

Given the equation of a parabola

y = 7x − x².

Differentiate the function

$\frac{dy}{dx}=7-2x$

For the point (1, 6), x = 1

Substitute x = 1 into the differential

$\frac{dy}{dx}=7-2(1)\\\frac{dy}{dx}=5$

Hence the slope is 5

Get the required equation:

Recall that $y-y_0 = m(x-x_0)$

$y-6 = 5(x-1)\\y-6=5x-5\\y=5x-5+6\\y=5x+1$

Hence the slope of the tangent line to the parabola at the point (1, 6) is 5

while the equation is y = 5x + 1

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