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Algebra Problem About Quadratic Function

Joystar1977

Active member
Jul 24, 2013
119
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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You are correct that the factored form:

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

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:

\(\displaystyle y=(x-2)^2-9\)

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 $(2,-9)$.
 

Joystar1977

Active member
Jul 24, 2013
119
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?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

\(\displaystyle y=(x-2)^2-9\)

is the easiest to use for graphing purposes. :D