Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
To determine the value of [tex]\( a \)[/tex] in the quadratic function's equation [tex]\( y = ax^2 + bx + c \)[/tex], we follow these steps:
1. Identify and list the points:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & -3 \\ \hline 1 & -3.75 \\ \hline 2 & -4 \\ \hline 3 & -3.75 \\ \hline 4 & -3 \\ \hline 5 & -1.75 \\ \hline \end{array} \][/tex]
2. Calculate the first differences:
[tex]\[ \begin{aligned} \Delta_1 &= -3.75 - (-3) = -0.75 \\ \Delta_2 &= -4 - (-3.75) = -0.25 \\ \Delta_3 &= -3.75 - (-4) = 0.25 \\ \Delta_4 &= -3 - (-3.75) = 0.75 \\ \Delta_5 &= -1.75 - (-3) = 1.25 \\ \end{aligned} \][/tex]
3. Calculate the second differences:
[tex]\[ \begin{aligned} \Delta^2_1 &= -0.25 - (-0.75) = 0.5 \\ \Delta^2_2 &= 0.25 - (-0.25) = 0.5 \\ \Delta^2_3 &= 0.75 - 0.25 = 0.5 \\ \Delta^2_4 &= 1.25 - 0.75 = 0.5 \\ \end{aligned} \][/tex]
You can observe that all the second differences are constant and equal to [tex]\( 0.5 \)[/tex].
4. Find the coefficient [tex]\( a \)[/tex] from the second differences:
- For a quadratic function [tex]\( y = ax^2 + bx + c \)[/tex], the second differences are equal to [tex]\( 2a \)[/tex].
- Given that the second differences are [tex]\( 0.5 \)[/tex], we can set up the equation:
[tex]\[ 2a = 0.5 \][/tex]
- Solving for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{0.5}{2} = 0.25 \][/tex]
Therefore, the value of [tex]\( a \)[/tex] in the function's equation is [tex]\( \boxed{\frac{1}{4}} \)[/tex]. Thus, the correct answer is [tex]\( \text{C.} \frac{1}{4} \)[/tex].
1. Identify and list the points:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & -3 \\ \hline 1 & -3.75 \\ \hline 2 & -4 \\ \hline 3 & -3.75 \\ \hline 4 & -3 \\ \hline 5 & -1.75 \\ \hline \end{array} \][/tex]
2. Calculate the first differences:
[tex]\[ \begin{aligned} \Delta_1 &= -3.75 - (-3) = -0.75 \\ \Delta_2 &= -4 - (-3.75) = -0.25 \\ \Delta_3 &= -3.75 - (-4) = 0.25 \\ \Delta_4 &= -3 - (-3.75) = 0.75 \\ \Delta_5 &= -1.75 - (-3) = 1.25 \\ \end{aligned} \][/tex]
3. Calculate the second differences:
[tex]\[ \begin{aligned} \Delta^2_1 &= -0.25 - (-0.75) = 0.5 \\ \Delta^2_2 &= 0.25 - (-0.25) = 0.5 \\ \Delta^2_3 &= 0.75 - 0.25 = 0.5 \\ \Delta^2_4 &= 1.25 - 0.75 = 0.5 \\ \end{aligned} \][/tex]
You can observe that all the second differences are constant and equal to [tex]\( 0.5 \)[/tex].
4. Find the coefficient [tex]\( a \)[/tex] from the second differences:
- For a quadratic function [tex]\( y = ax^2 + bx + c \)[/tex], the second differences are equal to [tex]\( 2a \)[/tex].
- Given that the second differences are [tex]\( 0.5 \)[/tex], we can set up the equation:
[tex]\[ 2a = 0.5 \][/tex]
- Solving for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{0.5}{2} = 0.25 \][/tex]
Therefore, the value of [tex]\( a \)[/tex] in the function's equation is [tex]\( \boxed{\frac{1}{4}} \)[/tex]. Thus, the correct answer is [tex]\( \text{C.} \frac{1}{4} \)[/tex].
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.