Question 199593: Hi all, I was hoping someone could help me with the following Graphing problems im having. I need to find the x and y intercepts, if any, and turning points for the following functions. (Each one on seperate graphs)
1. y = 2x^2-8x+9
2. y = 2x^2-3x-1
3. y = -2x^2-7x+4
4. y = x^2-x-2
I f you could please show the working with descriptions on each equations that would be very helpful. A fully labelled sketch of the function including the points and axis of symmetry would also be great.
Thanks, -Nick
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
It seems like you want someone to do all of your work for you. Not going to happen. What I will do is show you the process for doing all of them, so that you can do your own work.
First thing is that these are all quadratic functions of the form:
 = ax^2 + bx + c)
And all subsequent discussion will be related to the components of this form.
 -intercept
The -intercept is the point where the graph intersects the -axis. The  -coordinate of this point is obviously 0, and the -coordinate is the value of the function when  , specifically:
 = a(0)^2 + b(0) + c = c)
Therefore the -intercept is the point )
-intercept(s)
The -intercept(s) are the point(s) where the graph intersects the -axis. The -coordinates of the -intercepts correspond to the real zeros of the function; in other words the real number solutions or roots of the equation:
Recall the discriminant for a quadratic equation. If  , you have two -intercepts. If  , you have one -intercept (and the graph is tangent to the -axis at the vertex of the parabola). If  you have zero -intercepts.
The -coordinate of an -intercept is 0. To determine the -coordinates, set the function equal to 0 and solve for using any appropriate method (factoring, completing the square, or the quadratic formula).
Turning Points
Since these are all quadratic functions of the form defined above, all of the graphs are of parabolas. Therefore, there is one and only one turning point, and that is the vertex of the parabola. The -coordinate of the vertex of the graph of  = ax^2 + bx + c) is given by:
And the -coordinate of the vertex is the value of the function at the -coordinate of the vertex, i.e:
 = f\left(\frac{-b}{2a}\right) =a\left(\frac{-b}{2a}\right)^2 + b\left(\frac{-b}{2a}\right) + c)
Axis of Symmetry
The axis of symmetry for any function in this form is the vertical line described by the equation:  , that is to say 
Additional points
If you feel the need to determine additional points to help you sketch the true shape of the graph, then just select values for different than the values that you calculated above. Because of symmetry, the graph will have a point at  that has a -coordinate equal to the -coordinate of the -intercept. Also because of symmetry, if the point ) is a point on the graph, and if  , then the point at  will also have a function value (read -coordinate) of 
John
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