Question on quadratic and linear equation/graph

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Additional Information: The curve of y=(4-x)(x+5) cuts the x-axis at A and B. A straight line cuts the curve at A and at C. The x-coordinate of C is 2. Find

(a) the coordinates of point C,

(b) the equation of line AC,

(c) the area of Triangle ABC

I am stuck at the (a) part. I can't find a solution to get the "y" of point C. I don't see any clue too. What should I do to get "y" of point C?

Re: Question on quadratic and linear equation/graph

Point C is on the curve of equation y = (4-x)(x+5) so as you know the x co-ordinate you can simply sub it into the equation of the curve to find the y co-ordinate.

Re: Question on quadratic and linear equation/graph

The y co-ordinate of Point C is

$\displaystyle y=(4-x)(x+5)=f(2)$

when x=2

as C is on both the line and curve.

Re: Question on quadratic and linear equation/graph

Quote:

Originally Posted by

**Archie Meade** The y co-ordinate of Point C is

$\displaystyle y=(4-x)(x+5)=f(2)$

when x=2

as C is on both the line and curve.

Quote:

Originally Posted by

**Ross137** Point C is on the curve of equation y = (4-x)(x+5) so as you know the x co-ordinate you can simply sub it into the equation of the curve to find the y co-ordinate.

Thanks, sorry for the late reply, I got it. =)