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Remember that a graph will have a y-intercept at the point (0, f (0)) . Though, in this case we have to ignore x = 0 and thus this graph will never cross the y-axis. It does get extremely close to the y-axis, but it will never cross or touch it & thus no y-intercept.
Next, remember that we can determine where a graph will have x-intercepts by solving f ( x ) = 0 .
For rational functions this might seem like a mess to deal along with. Though, there is a nice fact regarding rational functions which we can use here. A rational function will be zero at a specific value of x only if the numerator is zero at that x & the denominator isn't zero at that x. In other terms, to determine if a rational function is ever zero all that we need to do is set the numerator equal to zero & solve. Once we have these solutions we just need to check that none of them make the denominator zero as well.
In our case the numerator is one and will never be zero and so this function will have no x- intercepts. Again, the graph will get extremely close to the x-axis but it will never touch or cross it.
At last, we have to address the fact that graph gets extremely close to the x and y-axis but never crosses. Since there isn't anything special about the axis themselves we'll use the fact that the x- axis is actually the line specified by y =0 and the y-axis is really the line specified by x = 0.
what is point slope form
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Suppose that a company has a fixed cost of $150 per day and a variable cost of x^2+x. Further suppose that the revenue function is R(x) = xp and the price per unit is given by p =
Determine which system below will produce infinitely many solutions. 2x + 5y = 24 2x + 5y = 42 3x - 2y = 15 6x + 5y = 11 4x - 3y = 9 -8x + 6y = -18 5x - 3y = 16 -2x + 3y =
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10x-25y+5=0
r-17
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