Respuesta :
[tex]h(w)=-0.39w^2\text{ -1.17b + 3}9[/tex]Explanation:
Quadratic equation is in the form:
[tex]h(x)=ax^2+bx\text{ + c }[/tex]when x = 2, h(x) = 0
[tex]\begin{gathered} 0=a(2)^2\text{ + b(2) + c} \\ 0\text{ = 4a + 2b + c} \\ \text{ 4a + 2b + c = 0 ...equation 1} \end{gathered}[/tex]when x = - 5, h(x) = 0
[tex]\begin{gathered} 0=a(-5)^2\text{ + b(-5) + c} \\ 0\text{ = 25a - 5b + c} \\ \text{25a - 5b + c = 0 }\ldots equation2 \end{gathered}[/tex]when x = 3, h(x) = -3.12
[tex]\begin{gathered} -3.12=a(3)^2+b(3)\text{ + c} \\ -3.12\text{ = 9a + 3b +c} \\ \text{9a + 3b +c = -3.12 }\ldots\text{equation 3} \end{gathered}[/tex]We need to find the values of a, b, c
subtract equation 2 from 1:
[tex]\begin{gathered} 4a\text{ - 25a +2b -(-5b) +c - c = }0\text{ - 0} \\ -21a\text{ + 2b + 5b + 0 = 0} \\ -21a\text{ + 7b = 0 ..equation 4} \end{gathered}[/tex]subtract equation 3 from 2:
[tex]\begin{gathered} 25a\text{ - 9a -5b - 3b + c - c = 0 -}(-3.12) \\ 16a\text{ - 8b + 0= 0}+3.12 \\ 16a\text{ - 8b = 3.12 ...equation 5} \end{gathered}[/tex]We would sove equation 4 and 5 to get a and b
Using elimination method:
multiply equation 4 by 8 and equation 5 by 7 to eliminate b
-168a + 56b = 0 ...equation 4
112a - 56b = 21.84 ...equation 5
Add both equations:
-168a +112a + 56b + (-56b) = 0+ 21.84
-56a + 56b - 56b = 21.84
-56a = 21.84
a = 21.84/-56
a = -0.39
substitute for a in any equation between 4 and 5:
112a - 56b = 21.84 ...equation 5
112(-0.39) - 56b = 21.84
-43.68 - 56b = 21.84
-56b = 43.68 + 21.84
b = 65.52/-56
b = -1.17
substittue for c in any of equation from 1-3:
25a - 5b + c = 0 ..equation 2
25(-0.39) - 5(-1.17) + c = 0
-9.75 + 5.85 + c = 0
-.39 + c = 0
c =3.9
When w = ?
h(w) = ?
replacing w with x in the formula for the quadratic above:
[tex]h(w)=aw^2\text{ + bw + c}[/tex][tex]\begin{gathered} h(w)=-0.39w^2\text{ + (-1.17)b + (3.9)} \\ h(w)=-0.39w^2\text{ -1.17b + 3}.9 \end{gathered}[/tex]