# Finding the equation of a parabola with 1 point

• Jaco Viljoen
In summary, the sketch on the previous page shows a parabola and a line intersecting at P and the origin O. The parabola is defined by the function f(x)=ax^2+bx+c, where Df=ℝ, Rf=[-2,∞), and b/(2a)=2. The line is defined by the function g(x)=-(1/2)x. The coordinates of R are (-2,-2) and the coordinates of P are (-5,2.5). The distance between P and Q is √{7.25}. To determine the maximum vertical distance between corresponding points on the graphs of f and g on the interval [xP,0], where xP
Jaco Viljoen

## Homework Statement

The sketch on the previous page shows the graph of a function f, which is a parabola with vertex R, and the graph of a function g, which is a straight line defined by g(x)=-(1/2)x. The graphs of f and g intersect at P and O(the origin). The function f is defined by:
f(x)=ax2+bx+c
Df=ℝ, Rf=[-2,∞) and b/(2a)=2.a)Find the coordinates of R.

## Homework Equations

g(x)=-(1/2)x
f
(x)=ax2+bx+c
Df=ℝ, Rf=[-2,∞) and b/(2a)=2.

## The Attempt at a Solution

if b/(2a)=2 then my x coordinate of the vertex = 2 correct? but on the graph it looks more like -2
Is this due to the missing (-)b

I am stumped on this question, I am sure It is quite obvious but I would appreciate a point in the right direction.
Thank you,
Jaco

#### Attachments

• Parabola.pdf
34 KB · Views: 268
Last edited:
If b/(2a) = 2, then b = 4a, so you can rewrite f(x) = ax^2 + 4ax + c. This has a minimum where 2a x +4a = 0, or where x=-2. So your original insight was right, but you might have tried to jump a logical step and missed where the negative belongs.

How can you tell what the y coordinate might be?

Jaco Viljoen
Hint: f(0) = 0.

##f(x) = ax^2 + bx + c = a(x-h)^2 + k##

Given: ##\frac{b}{2a} = 2##

The range of the function is from -2 to ##\infty##. What does this tell you about the y-coordinate of R?

Not only should you be able to find the coordinates of R. You should be able to find the equation for the parabola (and thus find any point on the parabola).

Jaco Viljoen
That is the picture:

Jaco Viljoen and SammyS
Rf={-2,∞)
so y=-2
so my vertex is (-2,-2) (Coordinates of R)
f(x)=1/2(x+2)^2-2

Last edited:
Jaco Viljoen said:
Rf={-2,∞)
so y=-2
so my vertex is (-2,-2) (Coordinates of R)
f(x)=1/2(x+2)^2-2
Yes, that's correct

Jaco Viljoen
Find:
a, b and c
y=1/2(x+2)^2-2
y=1/2(x+2)(x+2)-2
y=1/2(x^2+2x+2x+4)-2
y=1/2x^2+2x+0
so a=1/2, b=2 and c=0

Jaco Viljoen said:
Find:
a, b and c
y=1/2(x+2)^2-2
y=1/2(x+2)(x+2)-2
y=1/2(x^2+2x+2x+4)-2
y=1/2x^2+2x+0
so a=1/2, b=2 and c=0
Also is correct.

Jaco Viljoen
Thank you Sammy,

Find the equation of a line that is parallel to the straight line and passes through R.

y=-1/2x+0 point R(-2,-2)
y=mx+b
-2=-1/2x+b
-2=-1/2(-2)+b
-2=1+b
b=-3

y=-1/2x-3

EM_Guy
Find the coordinates of P
(1/2)x^2+2x+0=-(1/2)x+0
(1/2)x^2+2x+(1/2)x=0
(1/2)x^2+2&(1/2)x=0
(x^2)/2+(5x)/2=0
x^2+5x=0
x(x+5)=0

x=0 and x=-5

so (-5,?)

y=-1/2(x)
y=-1/2(-5)
y=2.5

so P (-5,2.5)

Last edited:
Jaco Viljoen said:
Find the coordinates of P
(1/2)x^2+2x+0=-(1/2)x+0
(1/2)x^2+2x+(1/2)x=0
Instead of dragging those 1/2 factors along, just multiply both sides of the equation by 2, which results in
##x^2 + 4x + x = 0##, or ##x^2 + 5x = 0##.

Jaco Viljoen said:
(1/2)x^2+2&(1/2)x=0
(x^2)/2+(5x)/2=0
x^2+5x=0
x(x+5)=0

x=0 and x=-5

so (-5,?)

y=-1/2(x)
y=-1/2(-5)
y=2.5

so P (-5,2.5)

SammyS and Jaco Viljoen
Thank you Mark,

Calculate the distance between P and Q
P(-5,2.5) Q(?,0)
Find Q

x intercepts:
y=(1/2)x^2+2x+0
((1/2)x+2)(x+0)
(1/2)x=-2 and x=0
x=-4 and x=0

Q(-4,0)

d=√{(x2-x1)2+(y2-y1)2}
d=√{(-4-(-5))2+(0-2.5)2}
d=√{(-4+5))2+(0-2.5)2}
d=√{12+(-2.5)2}
d=√{1+6.25}
d=√{7.25}

The distance between P and Q is √{7.25}

Last edited:
I do not understand the following question, some help please.

Calculate the maximum vertical distance between corresponding points on the graphs of f and g on the interval[xP,0], where xP denotes the x-coordinate of P.

Jaco Viljoen said:
I do not understand the following question, some help please.

Calculate the maximum vertical distance between corresponding points on the graphs of f and g on the interval[xP,0], where xP denotes the x-coordinate of P.
The vertical distance between the line and the parabola is ##h = -\frac x 2 - \frac{x^2}{2} - 2x##. This distance function turns out also to be a quadratic, so you can find its maximum value by techniques you already know.

SammyS said:
Yes, that's correct
hello Jaco and Sammy i have a question i can see on the equation you put 1/2 (x+2)^2-2 i am tryna find out where the 1/2 came from looking from the g=-1/2x does it mean when its parallel the -1/2 becomes positive 1/2

Hi Toni,
ƒ(g) is a line,
f(x)=1/2(x+2)^2-2 is the parabola.

so no.

I used:
ƒ(x)=a(x-h)^2+k to find a, 1/2(x+2)^2-2

Tony Mondi
Jaco Viljoen said:
Hi Toni,
ƒ(g) is a line,
f(x)=1/2(x+2)^2-2 is the parabola.

so no.

I used:
ƒ(x)=a(x-h)^2+k to find a, 1/2(x+2)^2-2
thanks i got it

Jaco Viljoen
how do I determine which function is greater than the other?

Last edited:
I am not sure I understand what is expected of me?
P is an intersection, so there is no difference in hight?

Surely the max vertical distance will be at the vertex or somewhere to the right?
Am I misunderstanding this?

Last edited:
Jaco Viljoen said:
how do I determine which function is greater than the other?
The graph you posted earlier clearly shows that the line is above the parabola for the interval in question.

Jaco Viljoen
Jaco Viljoen said:
I am not sure I understand what is expected of me?
P is an intersection, so there is no difference in hight?
Correct. The other intersection point is at the origin, so the y-coordinates of the line and the parabola are also equal at that point.
Jaco Viljoen said:
Surely the max vertical distance will be at the vertex or somewhere to the right?
Am I misunderstanding this?
I think the max. vertical distance is at some point to the left of the parabola's vertex. I worked it out several days ago, but have since thrown away my work, so I'm not absolutely certain of this. I gave a hint in post #17, so take a look at that one again.

Jaco Viljoen
g(x)-f(x)=d(x)
g(x)-f(x)
d(x)={(-1/2)x}-{(1/2)x^2+2x}
d(x)=(-1/2)x^2-2&(1/2)x

x=-b/2a
x=(-(2&1/2)/(2(-1/2))
x=-2.5

d(x)=(-1/2)x^2-2&(1/2)x
d(x)=(-1/2)(-2.5)^2-2&(1/2)(-2.5)
d(x)=3.125 maximum vertical distance

Last edited:
Mark44 said:
I think the max. vertical distance is at some point to the left of the parabola's vertex. I worked it out several days ago, but have since thrown away my work, so I'm not absolutely certain of this. I gave a hint in post #17, so take a look at that one again.

Hi Mark,
Does this look correct?

Jaco Viljoen said:
g(x)-f(x)=d(x)
g(x)-f(x)
d(x)={(-1/2)x}-{(1/2)x^2+2x}
d(x)=(-1/2)x^2-2&(1/2)x

x=-b/2a
x=(-(2&1/2)/(2(-1/2))
x=-2.5

d(x)=(-1/2)x^2-2&(1/2)x
d(x)=(-1/2)(-2.5)^2-2&(1/2)(-2.5)
d(x)=3.125 maximum vertical distance

Yes, this looks fine.

## 1. How do I find the equation of a parabola with 1 point?

To find the equation of a parabola with 1 point, you will need to use the general equation for a parabola, y = ax^2 + bx + c. Plug in the given coordinates of the point into this equation, and solve for a, b, and c using algebraic manipulation.

## 2. What is the general equation for a parabola?

The general equation for a parabola is y = ax^2 + bx + c, where a, b, and c are constants. This equation represents a parabola that opens either upwards or downwards, depending on the value of a.

## 3. Can I find the equation of a parabola with only one point and no other information?

No, you will also need to know the direction the parabola opens (up or down) in order to find the equation. This information can be provided through another given point, the vertex, or the axis of symmetry.

## 4. Is there a specific method or formula for finding the equation of a parabola with 1 point?

Yes, you can use the general equation for a parabola and plug in the coordinates of the given point. From there, you can use algebraic manipulation to solve for the values of a, b, and c.

## 5. Can I use the equation of a parabola with 1 point to graph the parabola?

Yes, once you have found the equation using the given point, you can plot this point on a coordinate plane and use the equation to find other points on the parabola. You can then connect these points to graph the parabola.

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