I'm not going to be a brat and tell you to start at the beginning.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.
Why introduce k? Besides you should define k. Is k any real number?[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]
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.
where did you get "f" from?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.
I did something wrongUsing 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?