MHB Algebra Problem About Quadratic Function

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The discussion centers on three equations representing the same quadratic function and how to determine their x-intercepts and graph them. The factored form, y = (x - 5)(x + 1), is identified as the easiest for finding x-intercepts due to the zero-factor property. For graphing, the vertex form, y = (x - 2)^2 - 9, is preferred as it clearly indicates the vertex at (2, -9) and shows the parabola's shape. Participants clarify that the graph should be a parabola, not a straight line, and emphasize the importance of using the correct forms for specific tasks. Overall, understanding the different forms of the quadratic function aids in both finding intercepts and graphing accurately.
Joystar77
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These 3 equations all describe the same quadratic function. What are the coordinates of the following points on the graph of the function? From which equation is each point most easily determined?

y = (x - 5) (x + 1)

y = x ^ 2 - 4x - 5

y = (x - 2) ^ 2 - 9

X-intercept, what are the points, equation, and explanation why that equation is the one

from which the x-intercepts are most easily determined?

Please tell me someone if this is correct or am I thinking of something else:

The simple way to graph y = (x - 5) (x + 1) is to generate at least four points, put those on graph paper and draw a straight line through them.

Here's how I generate the required points:

Use the equation, y = (x -5) (x + 1) and choose an integer for x, say x = 2, and substitute this into your equation to find the corresponding value of y.

y = (x - 5) (x + 1)

y = (2-5) (2 + 1)

y = (-3) (3)

y = -9

So, my first two points has coordinates of (2, -9). Now am I suppose to repeat this operation with a different value of x, say x = 4.

y = (x - 5) (x + 1)

y = (4 -5) (4 + 1)

y = (-1) (5)

y = -5

So, my second two points has coordinates of (4, -5).

Now mark these two locations on graph paper starting at the origin of my graph (where the x-axis crosses the y-axis), go to the right of 2 squares (x = 2) then down 9 squares
y = -9) and mark your first point.

For the second point, again, start at the origin and go right 4 squares (x = 4) and then down 5 squares (y = -5) and mark your second point.

Using a straight edge, draw a line joining these two points. You have now graphed the equation y = ( x -5) (x + 1).

Compare your graph with the graph of y = (x -5) (x + 1).

Am I starting this out right or am I thinking of something different? Please somebody let me know.
 
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You are correct that the factored form:

$$y=(x-5)(x+1)$$

is the easiest form for determining the $x$-intercepts. These intercepts occur for $y=0$, and this factored form allows us to use the zero-factor property to quickly determine them by equating each factor to zero and solving for $x$.

As far as graphing the function, you need to observe that it is a quadratic function, and so its graph will be that of a parabola, not a straight line. For graphing, the easiest way is to use the vertex form:

$$y=(x-2)^2-9$$

and observe that this is simply the graph of $y=x^2$ translated two units to the right and nine units down. Its vertex is at $(2,-9)$.
 
Why is the equation y = (x - 2) ^2 - 9 the one out of all three equations listed below:

y = (x - 5) (x + 1)

y = x^2 - 4x - 5

y = (x - 2) ^2 - 9

the one from which the x-intercepts are most easily determined? Can you explain this?
 
Joystar1977 said:
Why is the equation y = (x - 2) ^2 - 9 the one out of all three equations listed below:

y = (x - 5) (x + 1)

y = x^2 - 4x - 5

y = (x - 2) ^2 - 9

the one from which the x-intercepts are most easily determined? Can you explain this?

It is the factored form:

$y=(x-5)(x+1)$

from which the $x$-intercepts are most easily determined, the reason for which I explained in my post above. However, the vertex form:

$$y=(x-2)^2-9$$

is the easiest to use for graphing purposes. :D
 
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