Question 134725
I am completely lost on how to do these. Any help would be great.
It says:
Sketch the graph of the following. Label vertex, x-and y-intercepts.
f(x) = x^2 - x - 12
Here's how you can sketch a graph
f(x) = y so we have:
y = x^2 - x - 12
:
We will find the values of y by substituting -5 to +5 in the equation
A table of these value will be:
 x | y
-------
-5 |+18
-4 |+8
-3 |0; when y = 0, this is one of the x intercepts
-2 |-6
-1 |-10
 0 |-12; notice when x = 0, the value of y is 12, this is the y intercept
+1 |-12; it has reached a minimum and started back up again
+2 |-10
+3 |-6
+4 | 0; this is the 2nd x intercept
+5 |+8
Plot these values, it should look like this:
{{{ graph( 300, 200, -10, 10, -16, 10, x^2-x-12) }}}
:
Vertex=( , )
To find the vertex, first find the axis of symmetry. This can be found using the formula x = -b/(2a)
In this equation a=1; b=-1
x = {{{(-(-1))/(2*1)}}}
x = {{{1/2}}} or .5
Find the vertex by substituting 1/2 for x in the original equation:
y = .5^2 - .5 - 12
y= .25 - .5 - 12
y = -12.25
The x/y values for the vertex is: .5, -12.25, this is the minimum as you can see by the graph
:
x-intercepts=( , );( , 
x intercepts are the points which the graph crosses the x axis (y=0)
you can find them by solving the equation
x^2 - x - 12 = 0
Factors to
(x-4)(x+3) = 0
x = +4
x = -3; the x intercepts
:
y-intercept=( , )
As we said the y intercept occurs when x = 0, you see when substitute 0 for x in the equation; y = -12
:
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