how do i find the exact coordinates of the vertex of the parabola that passes through (1,3) (2,5) (4,4)?

Answer 1

First you find the equation of the parabola. But in this case, there is not enough information to describe a unique parabola. There are multiple solutions. It is customary in this type of problem to state the orientation of the parabola.

If the parabola is of the form

y = ax² + bx + c

Then we have three equations in 3 unknowns.

a + b + c = 3
4a + 2b + c = 5
16a + 4b + c = 4

Solving we get:

a = -5/6
b = 9/2
c = -2/3

y = (-5/6)x² + (9/2)x - 2/3

Take the derivative to find where the slope is zero.

dy/dx = -5/3x + 9/2 = 0
(5/3)x = 9/2
x = (9/2)(3/5) = 27/10 = 2.7

Plugging in we solve for y.

y = (-5/6)(27/10)² + (9/2)(27/10) - 2/3 = 649/120 ≈ 5.408

The vertex is at (27/10, 2107/120) ≈ (2.7, 5.408)
_______________________

But this problem can also be solved like this.

x = ay² + by + c

Again we have three equations in 3 unknowns.

9a + 3b + c = 1
25a + 5b + c = 2
16a + 4b + c = 4

Solving we get:

a = -5/2
b = 41/2
c = -38

x = (-5/2)y² + (41/2)y - 38

Take the derivative to find where the slope is zero.

dx/dy = -5x + 41/2 = 0
5y = 41/2
y = 41/10 = 4.1

Plugging in we solve for x.

x = (-5/2)(41/10)² + (41/2)(41/10) - 38 = 161/40 = 4.025

The vertex is at (41/10,161/40) = (4.1,4.025)

Answer 2

If parabola is of the form
y = ax^2 + bx + c,
thn just sub in the three points, you have three equations from which you can find a, b, c, then the vertex is x = -b/(2a), y = whatever you get when you sub the x value in the parabola equation.






See below




Sub (1,3)
3 = a + b + c

Sub (2,5)
5 = 4a + 2b + c

Sub (4,4)
4 = 16a + 4b + c

Subtract the first from the second, getting
3a + b = 2

Now the third from the second:
-12a - 2b = 1

To solve
3a + b = 2
-12a - 2b = 1
double the first and add to the second:
-6a = 5
Hence a = -5/6,
sub in 3a + b = 2 to get b = 9/2,
and sub both of these in a+b+c=3 to get
c = -2/3

Vertex is where x = -(9/2)/(-5/3) =2.7, y = 5+49/120 -- since the question says "exact" I guess you put it in this form rather than the decimal approximation of 5.40833

Answer 3

You have to know that the chord of a parabola is always parallel to the tangent at the point of middle abscissa. Here that gives you that the tangent at x=3/2 is 2 and at 3 is -1/2. Therefore the slope is zero for some x between 3/2 and 3, actually 4 times nearer from 3 than from 3/2 because the slope moves linearly with x. So the vertex has abscissa 3-(1/5)(3/2)=2.7. The slope at x=2 is 2+(1/3)(-5/2)=7/6. So the slope of the chord between the point of abscissa x=2 and the vertex is 7/12.
Hence the coordinates of the vertex are 2.7 and 2+ 0.7*7/12=289/120.

Answer 4

A parabola that opens up or down (down in this case - try plotting it to see it) has the general equation:

y = a(x-h)^2 + k where (h, k) is the vertex and a is negative

Plug in given points to get three equations and three unknowns:
(1, 3) => 3 = a(1-h)^2 + k (Equation 1)
(2, 5) => 5 = a(2-h)^2 + k (Equation 2)
(4, 4) => 4 = a(4-h)^2 + k (Equation 3)

From equation 1: k = 3 - a(1-h)^2

Substitute this into equations 2 and 3 to get:

5 = a(2-h)^2 + 3 - a(1-h)^2
4 = a(4-h)^2 + 3 - a(1-h)^2

Simplifying:

2 = a(4-4h+h^2-1+2h-h^2) = a(3-2h)
1 = a(16-8h+h^2-1+2h-h^2) = a(15-6h)

So we have the following system:

2 = 3a - 2ah
1 = 15a - 6ah

Multiplying the top equation by -3, we have:

-6 = -9a + 6ah
1 = 15a - 6ah

Adding the two equations we get:

-5 = 6a

So a = -5/6 (which is negative as needed)

Plugging this into 2 = 3a - 2ah OR 1 = 15a - 6ah:

2 = 3(-5/6) - 2(-5/6)h
2 = -5/2 + (5/3)h
9/2 = (5/3)h
27/10 = h

1 = 15(-5/6) - 6(-5/6)h
1 = -75/6+5h
81/6 = 5h
81/30 = h
27/10 = h

Finally, we plug a = -5/6 and h = 27/10 into k = 3 - a(1-h)^2

k = 3 - (-5/6)(1-(27/10))^2
= 3 + (5/6)(-17/10)^2
= 3 + (5/6)(289/100)
= 360/120 + 289/120
= 649/120

So our vertex is: (h, k) = (27/10, 649/120) = (2.7, 5.4)

Answer 5

You know it's a parabola, so it's going to have the form y = m*[(x-a)^2] + b. If you find out what a and b are, you'll know what the coordinates of this vertex are. Since you have three points (x,y), just write down three equations of this form and solve for m, a, and b.

If this is supposed to be an "easy" question, you can usually assume m=1.

Answer 6

let y=f(x)
and y=ax^2+bx+c where a,b,c are constants
(1,3) on the curve 3=a(1)^2+b(1)+c
a+b+c=3............(1)
(2,5)on the curve 5=a(2)^2+b(2)+c
4a+2b+c=5..........(2)
(4,4)on the curve 4=a(4)^2+b(4)+c
16a+4b+c=4............(3)
subtract(1)from (2)we get 3a+b=2.......(4) *-2
subt.(2)from(3)we get 12a+2b=-1......(5)
by solving (4) and(5) a=-5/6 b=9/2
from (1) c=8/3
then f(x)=-5/6x^2+9/2x+8/3
vertex x=(-9/2)/(2*-5/6)=27/10=2.7
y=f(2.7)=5/6(2.7)^2+9/2(2.7)+8/3=.......
then
vertex=(2.7,........)