What value of k will make 3x^2+kx)-4 an even function
The Subject is not 11th grade, but Math.
ra
I don't know about "even function," but to be factored, k = -1.
3x^2 - x - 4 = (3x - 4)(x + 1)
I hope this helps.
To determine the value of k that will make the quadratic function 3x^2 + kx - 4 an even function, we need to analyze the symmetry of the function with respect to the y-axis.
An even function is symmetric with respect to the y-axis, which means that if (a, b) is a point on the graph of the function, then (-a, b) should also be a point on the graph.
Let's substitute (-x) for x in the quadratic function and see if it remains the same:
f(x) = 3x^2 + kx - 4
f(-x) = 3(-x)^2 + k(-x) - 4
= 3x^2 - kx - 4
For f(x) to be an even function, f(-x) must be identical to f(x).
Comparing the two expressions, we can see that the coefficient of x in f(x) is kx, while the coefficient of x in f(-x) is -kx. For f(x) to be an even function, these two expressions must be equal.
Therefore, we have the equation kx = -kx.
To get kx on one side and -kx on the other side, we can add kx to both sides:
kx + kx = -kx + kx
2kx = 0
Now, we can solve for k by dividing both sides of the equation by 2x:
2kx / (2x) = 0 / (2x)
k = 0
Hence, the value of k that will make the quadratic function 3x^2 + kx - 4 an even function is k = 0.