# Math Help - Parabolas

1. ## Parabolas

How do I figure out problems similar to the following by graphing y = 2x2 + 3?

2. What are you trying to find..Also your equation $y = 2x2 + 3$, does it mean y=4+3 or y=4x+3 or...you just mistype and its suppose to be y=2x+3?

Edit
Never mind, I assume you mean $y=2x^2+3$ right? But the question still exists, what are you trying to find?

3. I mean how do I graph it as a parabola?

4. Find the y intercept, which means setting x=0, and the y intercept in this case is (0,3)
Find the x intercepts using the quadratic equation or factoring- There is no x-intercepts in that problem
Find the axis of symmetry, which is $x=-b/(2a)$ to find the min or the max otherwise known as vertex, in this case it would be the same as the y-intercept
Look at the sign to see if it's concave up or down
I think thats basically it to graph simple parabolas like the example you gave.

5. I'm sorry but I am still a bit confused... Is there a simpler way?

6. Easier way...plug in x values...or use the calculator
I assume you know what vertex, and setting x=0 and y=0 does since this is in the pre-calc section
The basic equation of a parabola is $y= ax^2+bx+c$
So to find the vertex of the equation you gave $y=2x^2+3$, you would do x=-b/2a, and in this cause there is no b, which means it is 0 so the x value of the vertex would be x=0/4=0 and plug in 0 for x in the equation you would get the vertex at (0,3).

Now that you have the vertex it's time to find the x intercept. To find the x-intercept you would either use the quadratic equation or factoring it if possible. The equation equation is:
Now solve for x. Whatever you get for x should the x intercepts..which I assume you know what it is. I don't think there is any x-intercepts in this case.
Now for the y intercept just set x=0 in the equation. Because that means when x=0, y=?. In this case when you set x=0, y will be 3 right?

7. I understand everything in this part...
Easier way...plug in x values...or use the calculator
I assume you know what vertex, and setting x=0 and y=0 does since this is in the pre-calc section
The basic equation of a parabola is
So to find the vertex of the equation you gave , you would do x=-b/2a, and in this cause there is no b, which means it is 0 so the x value of the vertex would be x=0/4=0
But I don't understand how you got the vertex....

8. -b/2a is the x value of the vertex because it is also the axis of symmetry for the parabola. To get the vertex, you plug in the x and you will get the y-value so the vertex would be (-b/2a, f(-b/2a)) with f(x) = y. If you are looking for where did I get -b/2a or how to derivative it, read this webpage

9. Is it any different to graph a linear equation such as y=2x^2+4? I don't know if this is a linear equation....

10. That is not a linear equation. It is a quadratic equation and no, it is not any different to solve $y=2x^2+4$

11. Can you please give an example of a linear equation and what makes it linear? thanks

12. Anything in the form of $y = mx + b$ where m is the slope and b is the y-intercept.

It is linear because when you graph it you do not get a curve. You get a straight line.

And to do this really easily just plug in x-values, from that you can get the y-values and then plot. Then do a simple connect the dots. You should get a pretty skinny curve with the vertex at $(0,3)$ You can find the vertex as linnus said by doing $\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$ You can use this only if it is in the form $ax^{2}+bx+c$