How do you identify the important parts of #y = 1/2(x-3)(x+1)# to graph it?

Answer 1

See stepwise explanation

#color(green)("Given: "y=1/2(x-3)(x+1).................................(1)#

#color(green)("Same as: "y=1/2(x^2-2x-3))#

#color(blue)(x^2 " is positive so" underline(" upwards U-shape")).................(2)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("To find "x_("vertex")" Consider the -2 of "-2x)#

#color(blue)(x_("vertex")=-1/2(-2)=+1)......................(3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("To find "y_("vertex")" Substitute (3) into (1)")#

#y=1/2(1-3)(1+1) = 1/2(-2)(2)=-2#

#color(blue)(y_("vertex")=-2)....................................(4)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("To find y-intercept")#

Substitute #x=0# in (1)

#y=1/2(0-3)(0+1) =1/2(-3)(1) = -3/2#

#color(blue)(y_("intercept") = -3/2)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("To find x-intercept")#

Substitute #y=0# in (1)

#0=1/2(x-3)(x+1)#

#(x-3)=0 -> x=+3#
#(x+1)=0-> x=-1#

#color(blue)(x_("intercept")-> (x ,y) -> (-1 , 0) " and "(+3 , 0))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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Answer 2

To graph the equation y = 1/2(x-3)(x+1), identify the key components:

  1. The vertex: The vertex of the parabola is at the point where the curve changes direction. In this case, the equation is in vertex form, so the vertex is at the point (h, k), where h is the x-coordinate of the vertex and k is the y-coordinate. To find the vertex, set the expressions inside the parentheses equal to zero and solve for x: (x - 3) = 0 and (x + 1) = 0. This gives x = 3 and x = -1. Substitute these values into the equation to find the corresponding y-values.

  2. The axis of symmetry: The axis of symmetry is a vertical line that passes through the vertex and divides the parabola into two symmetrical halves. In this case, the axis of symmetry is the vertical line passing through the x-coordinate of the vertex.

  3. The direction of opening: Since the leading coefficient (1/2) is positive, the parabola opens upwards.

  4. The x-intercepts: To find the x-intercepts, set y equal to zero and solve for x. These are the points where the parabola crosses the x-axis.

  5. The y-intercept: To find the y-intercept, set x equal to zero and solve for y. This is the point where the parabola crosses the y-axis.

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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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