Parabola given only two points!

[rebeccakcarroll]

Okay I really need help with this! I was asked to draw three different function on the same axes such that f(2)=1 and f(5)=3. I made the conclusion that my two points then are (2,1) and (5,3) and was able to make the function f(x)=(2/3)x-(1/3) which is a line and the function f(x)=l(2/3)x-(1/3)l which makes a v as all absolute value functions do. However, I am having trouble creating a simple parabola given the fact I only have two points that I know of. Any and all help appreciated!

 

[HallsofIvy]

There exist an infinite number of parabolas passing through two given points. Even if you require that the axis of symmetry be parallel to the y-axis any parabola of the form \(\displaystyle y= ax^2+ bx+ c\) has three coefficients to be determined so require three points to give three equations.

But in an equation of the form \(\displaystyle y= x^2+ bx+ c\) (taking the leading coefficient to be 1) we have only two coefficients to be determined. In order that the parabola pass through (2, 1) and (5, 3), we must have \(\displaystyle 1= 4+ 2b+ c\) and \(\displaystyle 3= 25+ 5a+ b\). Solve those two equations for a and b.

 

[John Marsh]

Don't Froget to check the image

Okay I really need help with this! I was asked to draw three different function on the same axes such that f(2)=1 and f(5)=3. I made the conclusion that my two points then are (2,1) and (5,3) and was able to make the function f(x)=(2/3)x-(1/3) which is a line and the function f(x)=l(2/3)x-(1/3)l which makes a v as all absolute value functions do. However, I am having trouble creating a simple parabola given the fact I only have two points that I know of. Any and all help appreciated!

The equation form is
(y-y0)^2=4a(x-x0)


Image -

http://www.wolframalpha.com/input/?i=parabola+%282%2C1%29+and+%285%2C3%29+​