[RESOLVED] solving equation

[.paul.]

I have an equation that I'm trying to solve:

y =- x²⁄16 + 21x⁄16 + 11⁄8

Multiply through by 16:
16y = -x² + 21x + 22

But that's as far as I can get.
I'd understand it if it was:

y = -x² + 21x + 22 = 0

But I'm not sure with 16y.

Can anyone explain how to simplify this equation?
Thanks for your help.

[jemidiah]

Well, if you want to solve your equation for x in terms of y, you'll have to use the quadratic formula, and there are two "branches" of solutions in general (the function is not invertible in general). Dividing by 16 is pretty superficial, but setting your quadratic equal to zero is a significantly simpler operation.

Anywho, treat y as a constant, so

-x²⁄16 + 21x⁄16 + (11⁄8 - y) = 0

Now apply the quadratic formula to get x = ...lots of ugly stuff, but with y's.... I can be more specific if needed.

[.paul.]

ok, but I can't see how to find a solution...

[jemidiah]

You wrote an expression, not an equation, so what do you mean by it "seems insoluble without a value for y"?

To take a simpler example, start with y=x^2. Doing the same operation, i.e. saying x^2 - y = 0 and applying the quadratic formula, gives x = +/- sqrt(y), exactly as it should. In this example, what more could you ask for?

I'm apparently not understanding you, so in general, what do you want to get out of whatever manipulations you're trying to perform?

[.paul.]

It was an assignment question that I answered incorrectly. I needed to solve x, but I was confused by y. In my answer I calculated x = 22 or x = -1
It must be possible because values for x were expected...

[jemidiah]

I think I'd have to see the question to know what you're trying to do. You again said you needed to "solve for x", and I've told you how to solve for x in terms of y. If you mean you need to solve for x when y=0, you can just set y=0 in the formula from post #3, which is the same as showing that -x²⁄16 + 21x⁄16 + 11⁄8 = 0 has solutions precisely at x=-1 and x=22 (it does). But then this really has nothing to do with y, so the first few lines of your first post don't make sense.

[.paul.]

(a) The graph of y = −x²/16 + 21x/16 + 11/8 is a parabola.
(ii) Use algebra to find the x-intercepts.

part (ii) I answered correctly. The part I didn't answer correctly was where I multiplied through by 16 incorrectly. Obviously if y did equal 0 then 16*0 also equals 0, but it's still an expression I don't understand

[jemidiah]

Well, y is 0 at the x-intercepts, since the x-intercepts are on the x-axis, and the x-axis is precisely the line consisting of points whose y-coordinate is 0.

Perhaps this will help? Consider the graphs of the following two functions:
y = −x²/16 + 21x/16 + 11/8
y = −x² + 21x + 22

(You can see them at Wolfram Alpha.)

These graphs have the same x-intercepts, since the first is obtained by "squashing" the second, namely scaling it down in the y-direction by a factor of 16.

Your first line "16y = -x² + 21x + 22" is true for all points (x, y) on the first graph, but at least to answer (ii) you'll set y=0 anyway and it won't matter. Not sure what more there is to say; good luck.

[.paul.]

That is helpful. Thanks again. I'll try running those graphs.
It's an assignment that I've completed and had returned, so i'm just working over some of my mistakes, but your input is appreciated...

[jemidiah]

Minor thing: I noticed a typo in my second graph's equation from post #8. I had written y = −x² + 21x + 11 instead of y = −x² + 21x + 22. This didn't affect the default Wolfram Alpha view of the graphs enough to really notice since they were "zoomed out" far enough. I've updated the link anyway, and the rest was correct.