Question

I'm given the equation A=-2x^2+40x and have to sketch a graph of A against x showing all relevant points.. I've worked out using the standard formulas: y=ax^2+bX+c and x=-b/2a that a=-2, b=40, c=0, x-coordinate = 10, y-coordinate = 200.. I'm very new to graphing parabolas so please help explain what to do next, and whether I've already made errors. Thanks :)

Answer 1

the easiest way to graph a parabola is to write its equation into its standard form.

can u do that?

Answer 2

Sorry, I'm not quite sure what you mean..

Answer 3

can u complete the square in the equation u r given?

Answer 4

I'm not quite sure how to do that either :(

Answer 5

\[ \large A=-2x^2+40x \]
\[ \large A\color{red}{-200}=-2(x^2-20x\color{red}{+100}) \]
\[ \large A-200=-2(x-10)^2 \]

Answer 6

this shows you that the vertex of the parabola is (10,200)
it opens downwards becuase of the -2 multiplying x^2
if u set A=0 u get
\[ \large 0=-2x^2+40x=-2x(x-20) \]
so x=0 or x=20 so the parabola crosses the x-axis at (0,0) and (20,0)

Answer 7

all u need to graph a parabola is three points: vertex and two other points (preferably the intercepts with x-axis)

Answer 8

the points have to be on opposite sides of the vertex.

Answer 9

So would the x-intercepts be drawn at y=0?

Answer 10

yes

Answer 11

good luck.
dinner time
sorry

Answer 12

Thanks so much for the help, I really appreciate it :)