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what is the discontinuity and zero of the function f(x)=3x^2+x-4/x-1; what are the zero(s) of the function f(x) = the quantity of 5 x squared minus 25 x, all over x; identify the vertical asymptotes of f(x) = 2 over quantity x squared plus 3 x minus 10; which graph represents the function f(x + 4); which graph represents the function f(x 2); which graph represents the function f(x 4x 2 4x 8 2x 2); which graph represents the function of f(x) = ?; identify the horizontal asymptote of f(x) = quantity 6 x minus 7 over quantity 11 x plus 8

Sagot :

For the function f(x) = 3x² + x - 4/(x-1), the discontinuity occurs at x = 1 and the solution for the expression are x = -1 and x = 2.

This is because the function is undefined at that point (the denominator of the fraction becomes 0). The zero(s) of the function can be found by setting the numerator (3x² + x - 4) equal to 0 and solving for x. This gives us the solutions x = -1 and x = 2.

For the function f(x) = (5x² - 25x)/x:

The zero(s) of the function can be found by setting the numerator (5x² - 25x) equal to 0 and solving for x. This gives us the solutions x = 0 and x = 5.

For the function f(x) = 2/(x² + 3x - 10):

The vertical asymptotes of the function occur at the values of x that make the denominator (x² + 3x - 10) equal to 0. Solving for x gives us the solutions x = -2 and x = 5.

For the function f(x+4):

The graph of this function can be obtained by taking the graph of y = f(x) and shifting it 4 units to the left.

For the function f(x²):

The graph of this function will be a parabola that opens upwards, with the vertex at the origin (0,0).

For the function f(x⁴ - 4x² + 4x - 8):

The graph of this function will be a quartic polynomial, with roots at x = -2, x = 0, x = 2, and x = 2.

It is not clear what the function f(x) = ? represents, as the question mark indicates that the function is not specified.

For the function f(x) = (6x - 7)/(11x + 8):

The horizontal asymptote of the function occurs at the value of y that the function approaches as x approaches infinity. To find this value, we can divide the leading terms of the numerator and denominator (6x and 11x, respectively). This gives us a horizontal asymptote at y = 6/11.

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