parabolas

[claudette]

when a quadratic equation as just one solution is it possible to graph its parabola and if so how do you do it?

 

[JeffM]

when a quadratic equation as just one solution is it possible to graph its parabola and if so how do you do it?

You graph it quite similarly to the way you do a straight line. Find the x and y intercepts and then pick a few other values of x and calculate the corresponding y values. Join the points so found with a smooth curve.

You can do this because the parabola is a very simple curve: you know its general shape before you even start. If the coefficient of the squared term is positive, the parabola is concave up. If the coefficient of the squared term is negative, the parabola is concave down. There will be one y intercept and zero, one, or two x intercepts depending on the sign of the quadratic's discriminant. (Zero if the discriminant is negative, one id the discriminant is zero, and two if the discriminant is positive.) In differential calculus, you will learn tools for drawing (or sketching) graphs of more complicated curves.

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Edit: You also need to find the co-ordinates of the vertex of the parabola.

Edit 2. Notice in serd's example that the coefficient of x^2 is positive and the parabola is concave up. Notice also in that example that the discriminant is

\(\displaystyle 0^2 - 4(2)(2) = 0 - 16 = -16 < 0.\) So there are no x-intercepts.

 

[serds]

Let's look at an example: y=2x²+2
for x=1 : 2*1²+2= 4
for x=2 : 2*2²+2= 10
for x=0 : 2*0²+2= 2
for x=-1 : 2*(-1)²+2= 4
for x=-2 : 2*(-2)²+2= 10

 

[claudette]

yes! thank you very much

You graph it quite similarly to the way you do a straight line. Find the x and y intercepts and then pick a few other values of x and calculate the corresponding y values. Join the points so found with a smooth curve.

You can do this because the parabola is a very simple curve: you know its general shape before you even start. If the coefficient of the squared term is positive, the parabola is concave up. If the coefficient of the squared term is negative, the parabola is concave down. There will be one y intercept and zero, one, or two x intercepts depending on the sign of the quadratic's discriminant. (Zero if the discriminant is negative, one id the discriminant is zero, and two if the discriminant is positive.) In differential calculus, you will learn tools for drawing (or sketching) graphs of more complicated curves.

Did this answer your question?

Edit: You also need to find the co-ordinates of the vertex of the parabola.

 

[HallsofIvy]

when a quadratic equation as just one solution is it possible to graph its parabola and if so how do you do it?

No, you cannot graph "equations", you graph functions!

The quadratic equation, \(\displaystyle x^2+ 6x+ 9= 0\) has just one solution. That is because \(\displaystyle x^2+ 6x+ 9\) is a "perfect square"- it is \(\displaystyle (x+ 3)^2\). Clearly it is never less than 0 and is 0 only at x= -3. If we let u= x+3, then \(\displaystyle y= x^2+ 6x+ 9= (x+ 3)^2= u^2\). If you know how to graph \(\displaystyle y= x^2\), the graph of \(\displaystyle y= x^2+ 6x+ 9\) is exactly the same except that its vertex is at (-3, 0) not (0, 0).

 

[lookagain]

when a quadratic equation has just one solution is it possible to graph its corresponding parabola and if so how do you do it?

No, you cannot graph "equations", you graph functions!

HallsofIvy, would it have helped if claudette had included the word "corresponding" above for emphasis?
She never stated, "is it possible to graph its (quadratic) equation?" So, it looks as if you were protesting something that isn't there.

 

[HallsofIvy]

Yes, rereading it, I agree. My apologies, Claudette.

 

[mmm4444bot]

when a quadratic equation [has] just one solution is it possible to graph its parabola

Yes. When there is only one solution value, then the vertex of the parabola is the x-intercept. In other words, the parabola touches the x-axis at only one spot, the vertex point.

You may find the vertex location, by using the formula for the x-coordinate of the vertex: -b/(2a)

After you plot the vertex point, plot the y-intercept.

Use symmetry to finish the graph. :cool:

By the way, serds' example has two solutions, and they are not Real. This is why the parabola in serds' example does not touch the x-axis; the roots of the polynomial are imaginary.