Graphing Quadratic

[Richard B]

The parabola [MATH]y=ax^2+bx+c [/MATH] has vertex [MATH](p,p)[/MATH] and [MATH]y[/MATH]-intercept [MATH](0,-p)[/MATH], where [MATH] p\neq 0 [/MATH]. What is [MATH]b[/MATH]?

I can't figure out where to start.

 

[Steven G]

In the point (0,-p), x = 0 and y = -p.
Plugging 0 of x and -p for y into the given equation can you figure out the value for c.

What is the formula to find the vertex of a parabola?

Please post back with your work.

 

[firemath]

The parabola [MATH]y=ax^2+bx+c [/MATH] has vertex [MATH](p,p)[/MATH] and [MATH]y[/MATH]-intercept [MATH](0,-p)[/MATH], where [MATH] p\neq 0 [/MATH]. What is [MATH]b[/MATH]?

I can't figure out where to start.

I'm not going to be a brat and tell you to start at the beginning.

[Jomo beat me to it]

 

[Richard B]

[MATH]y=ax+bx+c[/MATH]
[MATH]\downarrow[/MATH]
[MATH]y=a(x+\frac{b}{a}x+\frac{c-k}a)+k[/MATH]
making sure that
[MATH]\frac{b}{2a}=\sqrt{\frac{c-k}a}[/MATH]

 

[Steven G]

[MATH]y=ax+bx+c[/MATH]
[MATH]\downarrow[/MATH]
[MATH]y=a(x+\frac{b}{a}x+\frac{c-k}a)+k[/MATH]
making sure that
[MATH]\frac{b}{2a}=\sqrt{\frac{c-k}a}[/MATH]

Why introduce k? Besides you should define k. Is k any real number?

OK, I guess that you are trying to complete the square.

y= ax²+bx + c

y= a( x² + (b/a)x) + c

y = a( x² + (b/a)x + (b/(2a))²) + c - b²/(4a). You can continue from here.

I would however use the fact that if f(x) = ax²+bx + c, then the x-value of the vertex is x= -b/2a. But they are telling that this x-value is p and f(-b/2a) = p. Along with knowing the y intercept you can find b.

The thing is that you should take our advice and try it.

Please post back.

 

[MarkFL]

The parabola [MATH]y=ax^2+bx+c [/MATH] has vertex [MATH](p,p)[/MATH] and [MATH]y[/MATH]-intercept [MATH](0,-p)[/MATH], where [MATH] p\neq 0 [/MATH]. What is [MATH]b[/MATH]?

I can't figure out where to start.

Using the given information, we find it implies:

[MATH]p=ap^2+bp+c[/MATH]
[MATH]-p=c[/MATH]
This, in turn, implies:

[MATH]p=ap^2+bp-p[/MATH]
Since \(p\ne0\) we may divided by \(p\) and arrange as:

[MATH]ap=2-b[/MATH]
Now, given that the vertex lies on the axis of symmetry, we know:

[MATH]p=-\frac{b}{2a}\implies ap=-\frac{b}{2}[/MATH]
Can you proceed?

 

[Richard B]

I would however use the fact that if f(x) = ax²+bx + c, then the x-value of the vertex is x= -b/2a. But they are telling that this x-value is p and f(-b/2a) = p. Along with knowing the y intercept you can find b.

where did you get "f" from?

 

[Steven G]

y and f(x) are the same. I chose to use f(x) so later on I could say that f(-b/2a) = p.

 

[Richard B]

Using the given information, we find it implies:

[MATH]p=ap^2+bp+c[/MATH]
[MATH]-p=c[/MATH]
This, in turn, implies:

[MATH]p=ap^2+bp-p[/MATH]
Since \(p\ne0\) we may divided by \(p\) and arrange as:

[MATH]ap=2-b[/MATH]
Now, given that the vertex lies on the axis of symmetry, we know:

[MATH]p=-\frac{b}{2a}\implies ap=-\frac{b}{2}[/MATH]
Can you proceed?

I did something wrong

I got

[MATH]p=-\frac{b}{2a}[/MATH]
[MATH]-2ap=b[/MATH]
insert value for ap

[MATH]ap=-\frac{b}{2}[/MATH]
[MATH]-2(-\frac{b}{2})=b[/MATH]
[MATH]b=b[/MATH]
(yes, I know b=b isn't the answer they were looking for)

 

[Steven G]

ap = 2-b and ap = -b/2
Solve this for b

Yes, if you have one equation and you solve for b and then plug that value for b back into the equation you will not really get anywhere.

 

[Steven G]

Here is how I would do this problem.

The y intercept of any parabola in the form y = ax^2 + bx + c is always c. So c=-p.

The updated equation is now y = ax^2 + bx -p

The x value of the vertex is -b/2a. Now when x=-b/2a = p, so p = a(p)^2 + b(p) - p

Then ap^2 + bp =2p and -b/(2a) =p

Since p\(\displaystyle \neq\)0, ap^2 + bp = 2p becomes ap + b = 2.

Solve the system -b/(2a) =p and ap+b=2.

Then a(-b/(2a)) + b =2. -b/2 + b = 2. b/2 = 2 so b=4.

 

[Dr.Peterson]

I'd start by writing it in vertex form: y = a(x - p)^2 + p. Then set the y-intercept equal to -p: ap^2 + p = -p; solve for a, plug that in and expand to see what the linear term is.