Is there an equation for a parabola which has a vertex at (0,1) and goes through the point (1,3)
Hello,Originally Posted by nath_quam
if the vertex is $\displaystyle V(x_v \vert y_v)$ then you can use the vertex-form of the parabola:
$\displaystyle p(x)=a\cdot (x-x_v)^2+y_v$
Plug in the coordinates of the vertex:
$\displaystyle p(x)=a\cdot (x-0)^2+1$. Now is left only the coefficient a. Plug in the coordinates of the point (I've named it A):
$\displaystyle 3=a\cdot (1-0)^2+1$ and you'll get a = 2.
So the parabola is: $\displaystyle p(x)=2\cdot (x)^2+1$
I've attached a diagram.
Greetings
EB
nath_quamIs there an equation for a parabola which has a vertex at (0,1) and goes through the point (1,3)
Hi Nath_Quam: In short, yes, there are infinitely many such parabolas that meet the conditions given in your narrative. To begin, we write your parabola in functional notation, i.e., f(x) = ax^2 + bx + c. [Note: There is nothing compelling us to use f(x). We could just as legitimately have written g(x)=..., h(x)=..., or, simply y= ax^2 + bx + c. All are quadratic equations and, hence, have parabolic graphical representations].
Now, recall that any parabola has x-value: -b/2a. According to your given data, the vertex is at (1,3). Hence 1 =-b/2a, and we immediately observe b=0 in function, f(x) = ax^2 + bx + c. So, f(x) = ax^2+c.
Hello, nath_quam!
Assuming it is a "vertical" parabola, it has the form: .$\displaystyle y\:=\:ax^2 + bx + c$Is there an equation for a parabola which has a vertex at (0,1) and goes through the point (1,3)
The vertex of a parabola is at: $\displaystyle x = \frac{-b}{2a}$
Since this parabola has its vertex at $\displaystyle x = 0$, we see that $\displaystyle b = 0.$
. . The equation (so far) is: .$\displaystyle y\:=\:ax^2 + c$
Since $\displaystyle (0,1)$ is on the parabola: .$\displaystyle 1\:=\:a\cdot0^2 + c\quad\Rightarrow\quad c = 1$
. . The equation is: .$\displaystyle y\:=\:ax^2 + 1$
Since $\displaystyle (1,3)$ is on the parabola: .$\displaystyle 3\:=\:a\cdot1^2 + 1\quad\Rightarrow\quad a = 2$
. . Therefore, the equation is: .$\displaystyle \boxed{y \:=\:2x^2 + 1}$
Greetings Earboth:Originally Posted by earboth
Late last night, I began the process of addressing the given inquiry. After completing a few lines, however, circumstances arose such that I was unable to finish that which I had begun. I thought I had clicked the "cancel" icon but...
This morning, I awoke to read two well articulated, mathematically complete responses to the originator's query. Any contributory words I may have to offer now, would be an inefficient streatch of my stiff, early morning, pre-caffine enhanced cerebrum.
And that's that. Some few hours from now, I shall delete the initial post, as well as this clarification of the same.
Enjoy your day.
Regards,
Rich B.