2x^2 - 5x + 3 = 3x + k
2x^2 - 8x + (3-k) = 0
We want both roots of this to be the same. That is, it must be a perfect square.
2(x-2)^2 = 2x^2 - 8x + 8
So, we want 3-k=8 or k=-5
SO, 3x-5 intersects the parabola in exactly one point.
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If the line does not intersect the parabola, then f(x)-g(x) = 0 must have a negative discriminant.
The line intersects the parabola when
4x+k = -3x^2 - x + 4
3x^2 + 5x + k-4 = 0
5^2 - 4(3)(k-4) = 25 - 12k + 48 = 73 - 12k < 0
73-12k < 0
k > 73/12
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
determine the value(s) of k such that the linear function g(x)=4x+k does not intersect the parabola f(x)=-3x^2-x+4
4 answers
math.
Math is not fun
Our 2 equations:
y = 2x^2 - 5x + 3
y = 3x + k
2x^2 - 5x + 3 = 3x + k
2x^2 - 8x + 3x - k = 0
For only 1 point of intersection, the discriminant should equal 0.
b^2 - 4ac = 0, a = 2, b = -8, c = 3 - k
-8^2 - 4(2)(3-k) = 0
64 - 8(3-k) = 0
64 - 24 +8k = 0
40 + 8k = 0
40 = -8k
-5 = k
y = 2x^2 - 5x + 3
y = 3x + k
2x^2 - 5x + 3 = 3x + k
2x^2 - 8x + 3x - k = 0
For only 1 point of intersection, the discriminant should equal 0.
b^2 - 4ac = 0, a = 2, b = -8, c = 3 - k
-8^2 - 4(2)(3-k) = 0
64 - 8(3-k) = 0
64 - 24 +8k = 0
40 + 8k = 0
40 = -8k
-5 = k
1 answer