Question 1

Make sure that the expression under the radical is not negative.

Question 2

$\begin{gathered} y=ax^2+bx+c \\ \text{Vertex (}-2,5\text{)} \\ P(-3,2) \\ For\text{ the vertex} \\ 5=a(-2)^2+b(-2)+c \\ 5=4a-2b+c\text{ (1)} \\ \text{For P} \\ 2=a(-3)^2+b(-3)+c \\ 2=9a-3b+c\text{ (2)} \\ Vertex\text{ is the Max point of the function, hence } \\ y^(\prime)(-2)=0 \\ y^(\prime)=2ax+b \\ 0=2a(-2)+b \\ -4a+b=0\text{ (3)} \\ \text{System of equation} \\ 4a-2b+c=5\text{ (1)} \\ 9a-3b+c=2\text{ (2)} \\ -4a+b=0\text{ (3)} \\ \text{From (3)} \\ -4a+b=0\text{ } \\ 4a=b \\ U\sin g\text{ b in (1)} \\ 4a-2(4a)+c=5 \\ 4a-8a+c=5 \\ -4a+c=5 \\ U\sin g\text{ b in (2)} \\ 9a-3(4a)+c=2\text{ } \\ 9a-12a+c=2 \\ -3a+c=2 \\ \text{New equation system} \\ -4a+c=5\text{ }(4) \\ -3a+c=2\text{ (5)} \\ \text{From (4)} \\ \text{Solving c} \\ c=5+4a \\ U\sin g\text{ c in (5)} \\ -3a+5+4a=2 \\ -3a+4a=2-5 \\ a=-3 \\ With\text{ the value of a, c will be found} \\ c=5+4(-3) \\ c=5-12 \\ c=-7 \\ \text{Now, with a, b will found} \\ 4a=b \\ b=4(-3) \\ b=-12 \\ \text{Therefore} \\ \text{The quadr}aticfunctionisf(x)=-3x^2\text{-12x-7} \end{gathered}$

0 Comments

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

0 Comments

Please log in or register to add a comment.

Other Questions