Question

k is a quadratic function where k(3) = 0, k(- 2) = 0, and k (- 5) = -5.52. Find an algebraic equationfor k(w).

Answer 1

Answer:

$y(x)=-\frac{23}{100}x^2+\frac{23}{100}x+\frac{69}{50}$

Explanation: From the three points, we can find three equations with three unknowns, by using the standard form of a quadratic function:

$\begin{gathered} y(x)=ax^2+bx+c \\ y(3)=9a+3b+c=0 \\ y(-2)=4a-2b+c=0 \\ y(-5)=25a-5b+c=-5.52 \\ \\ \therefore\rightarrow \\ 9a+3b+c=0\Rightarrow(1) \\ 4a-2b+c=0\Rightarrow(2) \\ 25a-5b+c=-5.52\Rightarrow(3) \end{gathered}$

Solving (1) (2) and (3) gives use a b c: which is solved as follows:

$\begin{gathered} a=-\frac{23}{25} \\ b=-\frac{23}{5} \\ c=-\frac{138}{25} \\ \therefore\Rightarrow \\ y(x)=-\frac{23}{100}x^2+\frac{23}{100}x+\frac{69}{50} \end{gathered}$