Question

At what point on the parabola y = 3x^2 + 2x is the tangent line parallel to the line y = 10x −2?

Answer 1

Answer:

( $\frac{4}{3}$ , 8 )

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = 10x - 2 ← is in slope- intercept form

with slope m = 10

Parallel lines have equal slopes

then the tangent to the parabola with a slope of 10 is required.

the slope of the tangent at any point on the parabola is $\frac{dy}{dx}$

differentiate each term using the power rule

$\frac{d}{dx}$ (a$x^{n}$ ) = na$x^{n-1}$ , then

$\frac{dy}{dx}$ = 6x + 2

equating this to 10 gives

6x + 2 = 10 ( subtract 2 from both sides )

6x = 8 ( divide both sides by 6 )

x = $\frac{8}{6}$ = $\frac{4}{3}$

substitute this value into the equation of the parabola for corresponding y- coordinate.

y = 3($\frac{4}{3}$ )² + 2

   = (3 × $\frac{16}{9}$ ) + 2

   = $\frac{16}{3}$ + $\frac{8}{3}$

   = $\frac{24}{3}$

   = 8

the point on the parabola with tangent parallel to y = 10x - 2 is ( $\frac{4}{3}$ , 8 )