Question 52351
<pre><font size = 5><b>Find an equation for a Parabola with 
vertex (1,0) containing the point(4,18)

To do this you must know that  

1. The equation of a parabola with vertex (h,k) is

   y = a(x-h)² + k

2. If h, k, and a point (x<sub>1</sub>,y<sub>1</sub>) are known for a 
   parabola with the equation above in 1, then 
   the value of "a" can be found by substituting 
   x<sub>1</sub> for x and y<sub>1</sub> for y and then solving for "a".

So for your problrm (h,k) = (1,0), and 
(x<sub>1</sub>,y<sub>1</sub>) = (4,18), so we substitute 1 for h, 
0 for k, 4 for x and 18 for y in

    y = a(x-h)² + k 

   18 = a(4-1)² + 0
   18 = a(3)² 
   18 = 9a
    2 = a

Now rewrite the equation 

   y = a(x-h)² + k

with 2 for a, 1 for h and 0 for k, leaving 
x and y as variables:

   y = 2(x-1)² + 0

   y = 2(x-1)²

and if you like you may multiply that out 
and get

  y = 2x² - 4x + 2      
   
It's graph is
{{{ graph( 300, 300, -2, 20, -2, 20, 2x^2-4x+2, -18+sqrt(.2-(x-4)^2), 18+sqrt(.2-(x-4)^2) ) }}}
===================================================================

If lines 2x - y = 17 and 5x + y = 11 are 
graphed what is their intersection point?

Write them one under the other and add 
them vertically term-by-term and the y's
will cancel:

2x - y = 17
5x + y = 11
-----------
7x     = 28
     x = 4

Substitute 4 for x in either equation
I'll arbitrarily pick the second one:

  5x + y = 11  
5(4) + y = 11
  20 + y = 11
       y = -9

So they intersect at the point (4, -9)

Here are their graphs:
{{{ graph( 300, 300, -2, 18, -18, 2, 11-5x, 2x-17) }}}
=============================================

If f(x)=x2-x, what is f(-1)

Just substitute -1 for x in

  f(x) = x² - x

and simplify the right side

  f(-1) = (-1)² - (-1)
  f(-1) = 1 + 1
  f(-1) = 2

=================================================  

Find an equation that represents a Parabola 
on the xy-plane with vertex(-3,2)opening 
upwards?

Like the first problem we know it has equation

   y = a(x - h)² + k

with h = -3 and k = 2.  We also need to know
that if "a" is a positive number the parabola
will open upward and if "a" is a negative
number it will open downward. ("a" may not be
0). 

Since it doesn't matter what "a" equals, as 
long as it is positive, we'll just choose "a" 
to be the easiest positive value we can think 
of, which is 1. 

  y = 1[x - (-3)]² + 2
  y = 1(x + 3)² + 2
  y = (x + 3)² + 2
  y = x² + 6x + 9 + 2
  y = x² + 6x + 11

Here is its graph

 {{{ graph( 300, 300, -6, 5, -6, 5, x^2+6x+11) }}}

Edwin</pre></font></b>