Quadratic Function

[jinx24]

"Find a quadratic function that satisfies the given conditions."

"The axis of symmetry is the line x = 1. The y-intercept is 1. There is only one x-intercept."

I don't understand how there can only be one x-intercept when the axis of symmetry is x = 1.

How do I get started with this?

The only thing I have come up with h = 1.

 

[Unco]

A vertical parabola will have one x-intercept if its vertex sits on the x-axis.

Drawing a sketch is always a good place to start.

 

[galactus]

You know the y-intercept is 1, so you have a start with \(\displaystyle ax^{2}+bx+1\)

You know it has an axis of symmetry of x=1, so the vertex will have coordinate(1,?).

Since the vertex can be found by \(\displaystyle \frac{-b}{2a}\), you can use

\(\displaystyle \frac{-b}{2a}=1\).

Find values for a and b which will result in 1 and satisfy your requirements.

Do you have an idea what they are?. Only one x-intercept ought to be a hint as to the y value of the vertex. Once you know the vertex coordinates and a point the parabola passes through, you can find the standard equation. A graphing calculator would sure come in handy right about now.

 

[trist666]

I believe the quadratic equation would be y=(x-1)^2
Seeing as the yintercept is 1, the slope would be 1/1
If the axis of symmetry is x=1 and there is only 1 x-intercept, it would only make sense that the vertex of the parabola is on (1,0) making it (x-1)^2.
I hope this is right, but I'm not sure.

 

[galactus]

Yep. :lol: