SOLUTION: Determine the value of k such that g(x)= 3x + k intersects the quadratic function f(x)=2x^2-5x+3 at exactly one point.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Determine the value of k such that g(x)= 3x + k intersects the quadratic function f(x)=2x^2-5x+3 at exactly one point.      Log On


   

Question 914680: Determine the value of k such that g(x)= 3x + k intersects the quadratic function f(x)=2x^2-5x+3 at exactly one point.
Found 2 solutions by Fombitz, richard1234:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
At only 1 point means the line is tangent to the curve.
Since the slope of the line is fixed, find when the slope of the tangent line to the curve has a slope of m=3.
The slope of the tangent line is the value of the derivative.
df%2Fdx=4x-5
So then,
4x-5=3
4x=8
x=2
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Find the y-coordinate now.
When x=2, f%282%29=2%284%29-5%282%29%2B3
f%282%29=8-10%2B3
f%282%29=1
The line must have the same y-coordinate, so,
g%282%29=3%282%29%2Bk=1
6%2Bk=1
k=-5
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Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
g(x) and f(x) intersect at one point --> f(x) - g(x) = 2x^2 - 5x + 3 - 3x - k = 2x^2 - 8x + (3-k) has exactly one real solution.

The discriminant is 64 - 4(2)(3-k), setting to zero (so that the quadratic has one solution) gives us (4)(2)(3-k) = 64 --> k = -5.