Finding x- and y-intercepts of a parabola.

I am asked to use algebra to find the x and y intercepts from the following equation;

y = - __x^2 __+ 3x + 8

2

Note; - x^2 above means - x^2 and - 2, I can't get the minus sign in the right place.

Am I working along the right lines here;

2y = - 2x^2 + 3x + 8

2y = x^2 + 8

y = x^2 + 8 - 2

y = x^2 + 6

I am not really sure I have understood the question?

Re: Using algebra to find intercepts

Y intercept is where the graph of a function intersect y axis. X intercept is where the graph of the function intersects the x axis.

To find y intercept(s) of a function, substitue x = 0 and solve for y and you get y intercept.

Similarly, to find x intercept(s), substitute y=0 and solve for x.

Some times you get no solution (a contradiction), which means there is no x or y intercept for that function.

Finding x- and y-intercepts of a parabola.

OK so I think I am learning slowly without software(Rofl)

I am looking for the x and y intercepts using algebra, so here I go;

y = - __x^2 __+ 3x + 8

divide 2

as previous, the - is belonging to both the x^2 and the division 2

I think I first get rid of the division 2

2y = - 2x^2 + 3x + 8

y = x + 6

when x = 0

y = 6

Finding the x intercept

when y = 0

0 = - 2x^2 + 3x + 8

0 = -2x(x - 6x) + 8

(x - 3) or (x + 4)

So the intercepts are;

+ 3 and - 4

Not 100% confident but how am I doing

Thanks

David

Re: Finding x- and y-intercepts of a parabola.

Quote:

Originally Posted by

**David Green** OK so I think I am learning slowly without software(Rofl)

I am looking for the x and y intercepts using algebra, so here I go;

y = - __x^2 __+ 3x + 8

divide 2

as previous, the - is belonging to both the x^2 and the division 2

I think I first get rid of the division 2

2y = - 2x^2 + 3x + 8

y = x + 6

when x = 0

y = 6

Finding the x intercept

when y = 0

0 = - 2x^2 + 3x + 8

0 = -2x(x - 6x) + 8

(x - 3) or (x + 4)

So the intercepts are;

+ 3 and - 4

Not 100% confident but how am I doing

Thanks

David

If the equation is $\displaystyle y = \frac{-x^2}{2} + 3x + 8$ then the y-intercept is trivial: Substitute x = 0 and get y = 8. I cannot fathom how you get y = 6.

To get the x-intercepts, substitute y = 0:

$\displaystyle 0 = \frac{-x^2}{2} + 3x + 8$

Multiply both sides by 2: $\displaystyle 0 = -x^2 + 6x + 16$.

At this stage you can use the quadratic formula to solve for x. Alternatively:

Multiply both sides by -1: $\displaystyle 0 = x^2 - 6x - 16 \Rightarrow 0 = (x - 8)(x + 2)$. The solution for x is trivial.

Sorry to say this, but you need to get urgent help with basic algebra from your teacher.

Re: Finding x- and y-intercepts of a parabola.

Quote:

Originally Posted by

**David Green** OK so I think I am learning slowly without software(Rofl)

I am looking for the x and y intercepts using algebra, so here I go;

y = - __x^2 __+ 3x + 8

divide 2

Please **don't** do that. Either learn latex or, if you insist on plain text y = -(x^2)/2 + 3x + 8 or y = (-x^2+3x+8)/2

I'm not sure what you've done between

Quote:

y = - x^2 + 3x + 8

divide 2

and

Quote:

2y = - 2x^2 + 3x + 8

y = x + 6

Exactly how did your +8 become a +6?

Re: Finding x- and y-intercepts of a parabola.

Quote:

Originally Posted by

**David Green** I am asked to use algebra to find the x and y intercepts from the following equation;

y = - __x^2 __+ 3x + 8

2

Note; - x^2 above means - x^2 and - 2, I can't get the minus sign in the right place.

Am I working along the right lines here;

2y = - 2x^2 + 3x + 8

2y = x^2 + 8 Mr F says: Nothing here makes sense. What has happened to the 3x in the previous line ...?? Why is the coefficient of x^2 not -2 anymore. Why wasn't the 8 multiplied by 2 as well. Your effort is admirable but you have serious troubles which need to be addressed by your teacher in 1-on-1 help.

y = x^2 + 8 - 2

y = x^2 + 6

I am not really sure I have understood the question?

..