Question 1141582
please help me to find the equation of the graph which meets the x-axis at
-4 and 2, 

<pre>Something was left out because there are infinitely many different
equations of graphs which meets the x-axis at -4 and 2. The easiest one is

y = (x+4)(x-2), which when multiplied out becomes

y = x²+2x-8 which has the graph:

{{{graph(2000/7,400,-6,4,-10,4,(x+4)(x-2))}}}</pre>

that part a) and part b)the coordinates of the turning point.

<pre>The turning-point of the graph of {{{y=ax^2+bx+c}}} is the point

{{{(matrix(1,3,-b^""/(2a^""),",",(4ac-b^2)/(4a^"")))}}} 

For y = x²+2x-8, a=1. b=2, c=-8

{{{(matrix(1,3,-(2)^""/(2(1)^""),",",(4(1)(-8)-(2)^2)/(4(1)^"")))}}}

which works out to be (-1,-9)

------------------------</pre>
Question two, the sketch of the graph y= 2+ 2x -x^2 

{{{graph(240,400,-2,4,-6,4,2+2x-x^2)}}}

passing through points E and G, a) find the coordinates of E and G.

<pre>Again, something was left out, because there is no way we can know what 
points 
E and G are.</pre>

b) find the maximum value of y. Thanking you in advance.<pre>

The maximum value of a quadratic graph that opens downward and the minimum
value of one that opens upward is the y-coordinate of the turning point.

The turning-point of the graph of {{{y=ax^2+bx+c}}} is the point

{{{(matrix(1,3,-b^""/(2a^""),",",(4ac-b^2)/(4a^"")))}}}

Its y-coordinate {{{(4ac-b^2)/(4a^"")}}}.  To determine a, b, and c, we
rewrite y= 2+2x-x² in its standard order

y = -x²+2x+2, so a=-1, b=2, c=2 

{{{(4(-1)(2)-(2)^2)/(4(-1)^"")}}}

That works out to be maximum of 3, and you can see that the turning point
is the highest point, and it is even with 3 on the y-axis.

Edwin</pre>