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.
Total area
= 2(5 × 15) + (12 - 5 - 5)(15 - 8)
= 150 + 14
= 164 m²
Answer:
i think their were 17 do you have any answer choices
Answer:
Chris gets £3000
Step-by-step explanation:
A : B : C
1 : 3 : 4 = 8 (add all the parts together)
£6000 ÷ 8 = 750
so the multiplier is x 750
1 : 3 : 4
↩ x 750
750 : 2250 : 3000
Chris gets £3000
Answer:
1.65 x 10^-9
Step-by-step explanation:
0.00000000165 = 1.65 x 10^-9 <----scientific form
