Thread: sketching curves, roots problem

heres the question: sketch the curve of y=4-7x-2xsqrd
i get y=4, but to get the two x roots i factorise as normal, and get the solutions x=1/2 and x=+4
however the textbook says that the second root is actually -4...

why?

Originally Posted by chartsy

heres the question: sketch the curve of y=4-7x-2xsqrd
i get y=4, but to get the two x roots i factorise as normal, and get the solutions x=1/2 and x=+4
however the textbook says that the second root is actually -4...

why?

Please show all details of your working - all steps.

for easiness i will make this smiley face the sign for squaredNod)
sketch curve of y=4-7x-2x

1) a<0 (-2) so curve is inverted u shape
2) find where it crosses y axis:
subst x=o into eqn
y=4
3)factorise to find the two roots of the curve
0=-2x-7x+4
0=-2x-8x+x+4

-2x(x-4) 1(x-4)
(-2x+1) (x-4)=0

1)x-4=0
x=4
2)-2x+1=0
x=1/2

so x=1/2 and x=4

so why is it -4?

Originally Posted by chartsy

1) a<0 (-2) so curve is inverted u shape
2) find where it crosses y axis:
subst x=o into eqn
y=4
3)factorise to find the two roots of the curve
0=-2x^2-7x+4
0=-2x^2-8x+x+4

-2x(x-4) 1(x-4) ... should be -2x(x+4) + 1(x+4)
(-2x+1) (x-4)=0

1)x-4=0
x=4
2)-2x+1=0
x=1/2

so x=1/2 and x=4

so why is it -4?

...

...

next time, use the caret (^) to indicate an exponent.

hmmm thanks for that....need to be more focused when factorising then

Originally Posted by chartsy

hmmm thanks for that....need to be more focused when factorising then

If you had checked it by expanding, you would have seen the mistake immediately.