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Sagot :
To find where the two linear functions intersect, we need to determine the point [tex]\((x, y)\)[/tex] that satisfies both functions.
1. First, let's determine the equations of the two lines.
For the table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 26 \\ \hline -2 & 18 \\ \hline 0 & 10 \\ \hline 2 & 2 \\ \hline \end{array} \][/tex]
We need to find the slope [tex]\(m\)[/tex] and the y-intercept [tex]\(b\)[/tex] of the first line. We can calculate the slope between any two points, for example:
[tex]\[ m_1 = \frac{18 - 26}{-2 - (-4)} = \frac{18 - 26}{-2 + 4} = \frac{-8}{2} = -4 \][/tex]
Now, we use one of the points to find [tex]\(b_1\)[/tex]. Using the point [tex]\((0, 10)\)[/tex]:
[tex]\[ y = mx + b \implies 10 = -4(0) + b_1 \implies b_1 = 10 \][/tex]
So, the equation of the first line is:
[tex]\[ y = -4x + 10 \][/tex]
2. For the second table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 14 \\ \hline -2 & 8 \\ \hline 0 & 2 \\ \hline 2 & -4 \\ \hline \end{array} \][/tex]
Again, calculate the slope:
[tex]\[ m_2 = \frac{8 - 14}{-2 - (-4)} = \frac{8 - 14}{-2 + 4} = \frac{-6}{2} = -3 \][/tex]
Using the point [tex]\((0, 2)\)[/tex] to find [tex]\(b_2\)[/tex]:
[tex]\[ y = mx + b \implies 2 = -3(0) + b_2 \implies b_2 = 2 \][/tex]
So the equation of the second line is:
[tex]\[ y = -3x + 2 \][/tex]
3. Find the intersection of the two lines [tex]\(y = -4x + 10\)[/tex] and [tex]\(y = -3x + 2\)[/tex]:
Set the equations equal to each other to solve for [tex]\(x\)[/tex]:
[tex]\[ -4x + 10 = -3x + 2 \][/tex]
Rearrange to solve for [tex]\(x\)[/tex]:
[tex]\[ -4x + 3x = 2 - 10 \][/tex]
[tex]\[ -x = -8 \][/tex]
[tex]\[ x = 8 \][/tex]
4. Substitute [tex]\(x = 8\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]:
Using the first equation [tex]\(y = -4x + 10\)[/tex]:
[tex]\[ y = -4(8) + 10 \][/tex]
[tex]\[ y = -32 + 10 \][/tex]
[tex]\[ y = -22 \][/tex]
Thus, the point of intersection is [tex]\((8, -22)\)[/tex]. Therefore, the correct option is:
[tex]\[ \boxed{(8, -22)} \][/tex]
1. First, let's determine the equations of the two lines.
For the table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 26 \\ \hline -2 & 18 \\ \hline 0 & 10 \\ \hline 2 & 2 \\ \hline \end{array} \][/tex]
We need to find the slope [tex]\(m\)[/tex] and the y-intercept [tex]\(b\)[/tex] of the first line. We can calculate the slope between any two points, for example:
[tex]\[ m_1 = \frac{18 - 26}{-2 - (-4)} = \frac{18 - 26}{-2 + 4} = \frac{-8}{2} = -4 \][/tex]
Now, we use one of the points to find [tex]\(b_1\)[/tex]. Using the point [tex]\((0, 10)\)[/tex]:
[tex]\[ y = mx + b \implies 10 = -4(0) + b_1 \implies b_1 = 10 \][/tex]
So, the equation of the first line is:
[tex]\[ y = -4x + 10 \][/tex]
2. For the second table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 14 \\ \hline -2 & 8 \\ \hline 0 & 2 \\ \hline 2 & -4 \\ \hline \end{array} \][/tex]
Again, calculate the slope:
[tex]\[ m_2 = \frac{8 - 14}{-2 - (-4)} = \frac{8 - 14}{-2 + 4} = \frac{-6}{2} = -3 \][/tex]
Using the point [tex]\((0, 2)\)[/tex] to find [tex]\(b_2\)[/tex]:
[tex]\[ y = mx + b \implies 2 = -3(0) + b_2 \implies b_2 = 2 \][/tex]
So the equation of the second line is:
[tex]\[ y = -3x + 2 \][/tex]
3. Find the intersection of the two lines [tex]\(y = -4x + 10\)[/tex] and [tex]\(y = -3x + 2\)[/tex]:
Set the equations equal to each other to solve for [tex]\(x\)[/tex]:
[tex]\[ -4x + 10 = -3x + 2 \][/tex]
Rearrange to solve for [tex]\(x\)[/tex]:
[tex]\[ -4x + 3x = 2 - 10 \][/tex]
[tex]\[ -x = -8 \][/tex]
[tex]\[ x = 8 \][/tex]
4. Substitute [tex]\(x = 8\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]:
Using the first equation [tex]\(y = -4x + 10\)[/tex]:
[tex]\[ y = -4(8) + 10 \][/tex]
[tex]\[ y = -32 + 10 \][/tex]
[tex]\[ y = -22 \][/tex]
Thus, the point of intersection is [tex]\((8, -22)\)[/tex]. Therefore, the correct option is:
[tex]\[ \boxed{(8, -22)} \][/tex]
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