The tangent line to the graph of y=2x^2−3x+1 at the point P(a,b) is parallel to the line y=5x. This tangent line crosses the x-axis at (d,0). How do I find the value of d?

2 Answers
Jul 19, 2018

Find the Derivative

Explanation:

y' = 4x - 3 = 5

4x = 8

a = 2

b = 2*2^2 - 3 * 2 + 1 = 3

Tangent line is y = 5x + c and passes through P

b = 5a + c

3 = 5*2 + c => c = -7

y = 0 = 5 * d - 7 => d = 7/5

d=7/5

Explanation:

Since the point P(a, b) lies on the curve: y=2x^2-3x+1 hence it will satisfy the equation of curve as follows

b=2a^2-3a+1\ ...........(1)

The slope m of the tangent line parallel to the given line: y=5x will be equal to the slope of given line i.e.

m=5

Now, the equation of the tangent line with slope m=5 & drawn at P(a, b)

y-b=5(x-a)

y=5(x-a)+b\ ........(2)

substituting y=5(x-a)+b in the equation of curve: y=2x^2-3x+1, we get

5(x-a)+b=2x^2-3x+1

2x^2-8x+5a-b+1=0

Since, the line: y=5(x-a)+b is tangent to the given curve at a single point hence above equation will have equal real roots i.e. determinant b^2-4ac of above quadratic equation will be 0 as follows

(-8)^2-4(2)(5a-b+1)=0

b=5a-7\ ........(3)

equating (1) & (3), we get

2a^2-3a+1=5a-7

a^2-4a+4=0

(a-2)^2=0

a=2, 2

substituting a=2 in (2), we get

b=5(2)-7=3

substituting values of a=2 & b=3 in (2), we get equation of tangent

y=5(x-2)+3

y=5x-7

Since, above tangent intersects x-axis at (d, 0) hence this point will satisfy above equation

0=5d-7

d=7/5