Determine the value of k in y=kx^2-5x+2 that will result in the intersection of the line y=-3x+4 with the quadratic at
a) two points (1 mark)
b) one points (1 mark)
c) no point (1 mark)
you have the line intersecting when
kx^2-5x+2 = -3x+4
kx^2-2x-2 = 0
Now recall the discriminant, d=b^2-4ac
If d>0 there are two roots
If d=0, there is one root
If d<0 there are no real roots.
The discriminant of your equation is
4+8k
I expect you can determine when that is negative, zero, or positive.
To determine the values of k that result in different numbers of intersections between the line and the quadratic, we need to set the equation of the line equal to the quadratic equation and then solve for x. By analyzing the number of solutions for x, we can determine the number of points of intersection between the line and the quadratic.
a) Two Points of Intersection:
We set y = -3x + 4 equal to y = kx^2 - 5x + 2.
This gives us the equation: kx^2 - 5x + 2 = -3x + 4.
Rearrange the equation to get: kx^2 - 2x - 2 = 0.
To find two points of intersection, the quadratic equation must have two distinct solutions for x. We can use the discriminant to determine the number of solutions. The discriminant is given by Δ = b^2 - 4ac, where a, b, and c are the coefficients of x^2, x, and the constant term respectively.
In this case, a = k, b = -2, and c = -2.
For two solutions, Δ > 0. So, we have b^2 - 4ac > 0. Plugging in the values, (-2)^2 - 4(k)(-2) > 0.
Simplifying the inequality gives us: 4 + 8k > 0.
Subtracting 4 from both sides: 8k > -4.
Dividing both sides by 8, we get: k > -1/2.
Therefore, to have two points of intersection, the value of k must be greater than -1/2.
b) One Point of Intersection:
Similar to the previous case, we set y = -3x + 4 equal to y = kx^2 - 5x + 2.
This gives us the equation: kx^2 - 5x + 2 = -3x + 4.
Rearranging the equation gives us: kx^2 - 2x - 2 = 0.
For one point of intersection, the quadratic equation must have only one real solution, which means the discriminant Δ must be equal to 0. So, we have b^2 - 4ac = 0.
Substituting the values, (-2)^2 - 4(k)(-2) = 0.
Simplifying: 4 + 8k = 0.
Subtracting 4 from both sides: 8k = -4.
Dividing both sides by 8, we get: k = -1/2.
Therefore, to have only one point of intersection, the value of k must be equal to -1/2.
c) No Point of Intersection:
If the line and the quadratic do not intersect, it means they are either parallel or the quadratic does not cross the line. The slope of the line is -3, so for no point of intersection, the quadratic should not intersect the line. This occurs when the slope of the quadratic is also -3.
The slope of the quadratic function y = kx^2 - 5x + 2 is given by the derivative dy/dx. Taking the derivative of the quadratic, we get dy/dx = 2kx - 5.
Setting the slope of the quadratic equal to the slope of the line (-3), we have 2kx - 5 = -3.
Rearranging the equation gives us: 2kx = -3 + 5.
Simplifying: 2kx = 2.
Dividing both sides by 2x (assuming x ≠ 0), we get: k = 1/x.
Therefore, for the quadratic and the line to have no point of intersection, k should be equal to 1/x, where x is any real number except 0.