A possible graph transformation f(kx) when 0 < k < 1 is?
1 answer:
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Answer:
a) 4x -3y ≥ -24
b) 2x +3y ≥ -12
c) 4x -3y ≤ 12
d) 2x +3y ≤ 24
Step-by-step explanation:
All of the given points are on the axes, so it is useful to know the intercept form of the equation for a line is ...
x/a +y/b = 1 . . . . . . where a is the x-intercept and b is the y-intercept.
<h3>a)</h3>The boundary line for the first side is ...
x/-6 +y/8 = 1
4x -3y = -24 . . . . . . multiply by -24 to put in standard form
The graph shows the quadrilateral is to the right of this line, so its inequality is ...
4x -3y ≥ -24
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<h3>b)</h3>The boundary line for side 2 has equation ...
x/-6 +y/-4 = 1
2x +3y = -12 . . . . . multiply by -12
The quadrilateral is to the right of this line, as well, so its inequality is ...
2x +3y ≥ -12
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<h3>c)</h3>We can see from the graph that the x-intercept of the boundary line for side 3 is x=3. The equation for that boundary line is ...
x/3 +y/-4 = 1
4x -3y = 12
The quadrilateral is to the left of this line, so its inequality is ...
4x -3y ≤ 12
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<h3>d)</h3>Apparently, side 4 is intended to use points already given. We assume it passes through (6, 4) and (0, 8). This boundary is parallel to side 2, so will have the same equation, but with a different constant.
2x +3y = 2(0) +3(8) = 24
The quadrilateral is to the left of this line, so its inequality is ...
2x +3y ≤ 24
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The second attachment shows a graph of the four inequalities with their shading.
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<em>Additional comment</em>
All of the inequalities are written in standard form, so the x-coefficient is positive for all of them. The direction of the inequality symbol can be chosen to select values of x that are greater than or less than those on the boundary line. If the shading is to be to the right, then larger x-values are required, and the inequality will have the form ax + by ≥ c (a > 0). If shading is to the left, then smaller x-values are required, and the inequality symbol points in the other direction.
The figure is a parallelogram, not a rectangle.
