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Sagot :
The Solution:
Given:
[tex]y=7x^2+bx+c[/tex]Given that the points: (-1,4) and (2,13) are on the graph of the given equation,
We are required to find the values of a and b.
Substitute (x= -1, y = 4) in the equation, we get:
[tex]\begin{gathered} 4=7(-1)^2+b(-1)+c \\ 4=7-b+c \\ -3=-b+c...eqn(1) \end{gathered}[/tex]Substitute (x= 2, y = 13) in the equation, we get:
[tex]\begin{gathered} 13=7(2)^2+b(2)+c \\ 13=28+2b+c \\ -15=2b+c...eqn(2) \end{gathered}[/tex]Solving eqn(1) and eqn(2) simultaneously by the elimination method:
Subtract eqn(1) from eqn(2):
[tex]\begin{gathered} -15--3=2b--b+c-c \\ -12=3b \end{gathered}[/tex]Divide both sides by 3.
[tex]b=\frac{-12}{3}=-4[/tex]Substitute -6 for b in eqn(1).
[tex]\begin{gathered} -3=-b+c \\ -3=-(-4)+c \\ \\ -3=4+c \\ -3-4=c \\ -7=c \\ c=-7 \end{gathered}[/tex]Therefore, the correct answers are:
b = -4
c = -7
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