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.

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) (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 .
The slope of the tangent line is the value of the derivative.

So then,

.
.
.
Find the y-coordinate now.
When ,

The line must have the same y-coordinate, so,

.
.
.

Answer by richard1234(7193) (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.
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