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

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

Math is not fun

math.

Ah, playing matchmaker between functions, I see! Let's find the values you're looking for:

For g(x) = 3x + k to intersect f(x) = 2x^2 - 5x + 3 at exactly one point, the discriminant of the quadratic equation formed by setting them equal to each other must be equal to zero. The discriminant is given by b^2 - 4ac. In this case, a = 2, b = -5, and c = 3 - k.

So, setting up the equation, we have (-5)^2 - 4(2)(3 - k) = 0.
Simplifying, 25 - 24(3 - k) = 0.
Expanding and rearranging, 25 - 72 + 24k = 0.
Combining like terms, 24k - 47 = 0.
Finally, isolating k, we get k = 47/24.

For g(x) = 4x + k to not intersect f(x) = -3x^2 - x + 4 at any point, we need the quadratic function to have no real solutions. In other words, its discriminant must be negative. Using a similar process as before, the discriminant is given by (-1)^2 - 4(-3)(4 - k).

So, we set up the equation (-1)^2 - 4(-3)(4 - k) < 0.
Simplifying, 1 + 12(4 - k) < 0.
Expanding and rearranging, 1 + 48 - 12k < 0.
Combining like terms, -12k + 49 < 0.
Finally, isolating k, we get k < 49/12.

So, for g(x) = 4x + k to not intersect f(x) = -3x^2 - x + 4, the value of k must be less than 49/12.

Hope this helps, and may your functions find the perfect match or lack thereof!

To determine the value of k such that the linear function g(x) = 3x + k intersects the quadratic function f(x) = 2x^2 - 5x + 3 at exactly one point, we need to find the condition that results in a single intersection point.

When two functions intersect, they have the same value at that point. Therefore, we can set g(x) equal to f(x) and solve for x:

3x + k = 2x^2 - 5x + 3

Rearranging the equation to put it in quadratic form:

2x^2 - 8x + (3 - k) = 0

For there to be exactly one intersection point, the quadratic equation must have a single root, which means the discriminant (b^2 - 4ac) must be equal to zero.

The discriminant is given by:

b^2 - 4ac = (-8)^2 - 4(2)(3 - k)

Simplifying:

64 - 24(3 - k) = 0

64 - 72 + 24k = 0

24k - 8 = 0

24k = 8

k = 8/24

k = 1/3

Therefore, the value of k that results in a single intersection point between g(x) and f(x) is k = 1/3.

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