Finding the equation of a parabola with 1 point

[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

 

[RUber]

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?

 

[RUber]

Hint: f(0) = 0.

 

[EM_Guy]

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

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?

http://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html

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).

 

[ehild]

That is the picture:

 

[Jaco Viljoen]

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

 

[SammyS]

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

 

[SammyS]

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,

 

[Jaco Viljoen]

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

 

[Jaco Viljoen]

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)

 

[Mark44]

Your answer below is correct. I've added some comments about how you could work more efficiently.

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 $x^2 + 5x = 0$.

(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)

 

[Jaco Viljoen]

Thank you Mark,

 

[Jaco Viljoen]

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}

 

[Jaco Viljoen]

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[x_P,0], where x_P denotes the x-coordinate of P.

 

[Mark44]

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[x_P,0], where x_P 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.

 

[Tony Mondi]

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

 

[Jaco Viljoen]

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]

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?

 

[Jaco Viljoen]

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

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

 

[Mark44]

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.

 

[Mark44]

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

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.

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

 

[Jaco Viljoen]

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?

 

[Mark44]

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.