Class 9th math 3.3 solution english PCTB

Question 7: Consider the function defined by \(f(x)=cx^2+d\), where \(c\) and \(d\) are constant numbers. If \(f(1)=6\) and \(f(-2)=10\), then find the values of \(c\) and \(d\).

Solution:

\(f(x)=cx^2+d\)

\(f(1)=c{(1)}^2+d\)

\(c+d=6\)\(\ \_\_\_\_\_\_\ (1)\)\(\quad \because f(1)=6\)

\(f(-2)=c{(-2)}^2+d\)

\(4c+d=10\)\(\ \_\_\_\_\_\_\ (2)\)\(\quad \because f(-2)=10\)

Subtracting equation \((1)\) from \((2)\), we have

\(\ \ 4c+d=10\)

\(\underline{\underset{-}{}\ \ c\underset{-}{+}d=\underset{-}{}6}\)

\(\ \ 3c\ \ \ \ \ \ \ =4\)

\(c=\large{\frac{4}{3}}\)

Put value of \(c\) in equation \((1)\), we have

\(\large{\frac{4}{3}}+d=6\)

\(d=6-\large{\frac{4}{3}}\)

\(d=\large{\frac{18-4}{3}}\)

\(b=\large{\frac{14}{3}}\)

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