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#?

Question

Christopher Agresta · Accepted Answer

Find the Derivative

#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#

Christopher Allers · Answer

#d=7/5#

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#