How do you graph #(6x^2 + 10x - 3) /( 2x + 2)#?

Answer 1

To graph the function (6x^2 + 10x - 3) / (2x + 2), we can follow these steps:

  1. Determine the domain of the function by finding the values of x for which the denominator (2x + 2) is equal to zero. In this case, the denominator is zero when x = -1. Therefore, the domain of the function is all real numbers except x = -1.

  2. Simplify the function by performing the division. Divide each term of the numerator (6x^2 + 10x - 3) by the denominator (2x + 2). This will give you the simplified form of the function.

  3. Once you have the simplified form, you can proceed to graph it. Plot points on the graph by choosing different values of x and calculating the corresponding y-values using the simplified function. You can choose values of x that are greater than, less than, and equal to -1 to get a sense of the behavior of the function around that point.

  4. Connect the plotted points with a smooth curve to obtain the graph of the function.

Note: It is important to remember that the graph may have a vertical asymptote at x = -1 due to the excluded value from the domain.

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Answer 2

When plotting graphs, I usually find the axis intercept points, the asymptotes, the stationary points and the points of inflection.

# f(x) = (6x^2+10x-3)/(2x+2) = 3x+2-7/(2x+2) #
To find the axis intercept points solve # f(x) = 0 # and # f(0) #: # f(x) = 0 # # 6x^2+10x-3 = 0 # # x = (-5 pm sqrt(43))/6 #
# f(0) = -3/2 #
Vertical asymptotes (denominator of f(x) = 0): # 2x+2 = 0 # # x = -1 # # lim_(x rarr -1^(+-)) f(x) = ""_+^(-) oo #
No horizontal asymptotes since: # lim_(x rarr pm oo} f(x) = pm oo #
However: # g(x) = 3x+2 # is an asymptote since # lim_(x rarr pm oo) -7/(2x+2) = 0^(""^(""_+^-)) #
# lim_(x rarr pm oo} f(x) rArr lim_(x rarr pm oo} g(x)^(""^(""_+^-)) #
Stationary points (first derivative is equal to zero): # (df)/dx ne 0, x in RR # So there are no stationary points.
Points of inflection (second derivative is equal to zero): # (d^2f)/dx^2 ne 0, x in RR # So there are no points of inflection.

graph{(6x^2+10x-3)/(2x+2) [-5, 5, -25, 25]}

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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