Question

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`?

Answer 1

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`

Answer 2

`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`