Given three points
P1(-2,8)
P2(0,-4)
P3(4,68)
We need the quadratic equation that passes through all three points.
Solution:
We first assume the final equation to be
f(x)=ax^2+bx+c .............................(0)
Observations:
1. Points are not symmetric, so cannot find vertex visually.
2. Using the point (0,-4) we substitute x=0 into f(x) to get
f(0)=0+0+c=-4, hence c=-4.
3. We will use the two other points (P1 & P3) to set up a system of two equations to find a and b.
f(-2)=a(-2)^2+b(-2)-4=8 => 4a-2b-4=8.................(1)
f(4)=a(4^2)+b(4)-4=68 => 16a+4b-4=68.............(2)
4. Solve system
2(1)+(2) => 24a+0b-12=84 => 24a=96 => a=96/24 => a=4 ......(3)
substitute (3) in (2) => 16(4)+4b-4=68 => b=8/4 => b=2 ..........(4)
5. Put values c=-4, a=4, b=2 into equation (0) to get
f(x)=4x^2+2x-4
Check:
f(-2)=4((-2)^2)+2(-2)-4=16-4-4=8
f(0)=0+0-4 = -4
f(4)=4(4^2)+2(4)-4=64+8-4=68
So all consistent, => solution ok.
If you mean standard 8.5x11 paper then 3 papers wide would make it 25.5in or 2.13 feet.
Answer: y = 2x + 50
Step-by-step explanation:
first find the slope (m):
slope = rise/run = y1-y2/x1-x2
(1,52); (2,54)
slope = (54-52)/(2-1) = 2/1 = 2; m = 2
now since we have the slope and a point, we will put it into point-slope form:
y = m(x - x1) + y1
y = 2(x - 3) + 56
y = 2x - 6 + 56
y = 2x + 50
Parrelel why is yes help me pro
