Write the equation in standard form using integers
(no fractions or decimals): 𝑦 = −2/3𝑥 − 1
6. Write an equation of the line that passes through
(2, -1) and is parallel to the graph of y = 5x – 2.
Write your final equation in slope-intercept form.
7. Write an equation of the line that passes through
(3, 5) and is perpendicular to the graph of y = –3x
+7. Write your final equation in slope-intercept
form.
8. Sketch a graph y = |x – 3| – 2 and describe the
translations.
6. Okay, I'll give it a shot! So, we know that the line is parallel to the graph of y = 5x - 2. That means the slopes of the two lines are equal. The slope of the given line is 5. Since parallel lines have the same slope, the slope of our new line will also be 5. Now, we just need to find the y-intercept. We know the line passes through (2, -1), so we can substitute these values into the slope-intercept form y = mx + b. We get -1 = 5(2) + b. Solving for b, we find b = -1 - 10. Simplifying that, we get b = -11. Putting it all together, the equation of the line is y = 5x - 11 in slope-intercept form.
7. Alrighty, let's get perpendicular! The given line is y = -3x + 7, which means the slope of our new line will be the negative reciprocal of -3, which is 1/3. Now, using the point (3, 5) that the line passes through, we can form the equation using slope-intercept form again: y - 5 = (1/3)(x - 3). Expanding that, we get y - 5 = (1/3)x - 1. Adding 5 to both sides, we end up with y = (1/3)x + 4. So, our final equation is y = (1/3)x + 4.
8. Ah, a graph question! Excellent! The equation y = |x - 3| - 2 describes an absolute value function. The absolute value symbol |x - 3| means that the value within it will always be positive or zero. So, if we rearrange the equation a bit, we can see that it's simply the distance between x and 3, minus 2. When x is greater than 3, the distance is positive, and when x is less than 3, the distance is negative. By subtracting 2, we shift the entire graph down by 2 units. So, the graph of y = |x - 3| - 2 is a "V" shape, centered at x = 3, and shifted down by 2 units.
6. To find the equation of a line that is parallel to the graph of y = 5x - 2, we need to find the line's slope. Since parallel lines have the same slope, the slope of the new line will also be 5.
We are given that the new line passes through the point (2, -1). Using the point-slope form of a linear equation, we can write the equation as:
y - y1 = m(x - x1)
Plugging in the values, we get:
y - (-1) = 5(x - 2)
y + 1 = 5x - 10
To write the equation in slope-intercept form (y = mx + b), we can isolate y:
y = 5x - 10 - 1
y = 5x - 11
So the final equation in slope-intercept form is y = 5x - 11.
7. To find the equation of a line that is perpendicular to the graph of y = -3x + 7, we need to find the negative reciprocal of the slope. The slope of the new line will be 1/3.
We are given that the new line passes through the point (3, 5). Using the point-slope form, we can write the equation as:
y - y1 = m(x - x1)
Plugging in the values, we get:
y - 5 = 1/3(x - 3)
To write the equation in slope-intercept form (y = mx + b), we can isolate y:
y = 1/3(x - 3) + 5
y = 1/3x - 1 + 5
y = 1/3x + 4
So the final equation in slope-intercept form is y = 1/3x + 4.
8. The equation y = |x - 3| - 2 represents an absolute value function. The graph of this equation is a V-shaped graph centered at the point (3, -2).
When x is less than 3, the expression inside the absolute value becomes negative, so y = -(x - 3) - 2 = -x + 5.
When x is greater than 3, the expression inside the absolute value becomes positive, so y = x - 3 - 2 = x - 5.
So the graph is a V-shape with the vertex at (3, -2), and the two arms of the V extending upwards and downwards from the vertex.
To write an equation in standard form using integers, we need to eliminate all fractions or decimals.
6. To find the equation of a line parallel to y = 5x - 2 and passing through (2, -1), we know that parallel lines have the same slope. So, the slope of the new line will also be 5. Using the point-slope form of a line, the equation will be y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Using the values (2, -1) and the slope 5, the equation becomes: y - (-1) = 5(x - 2). Simplifying, we get y + 1 = 5x - 10.
To write it in slope-intercept form, we can isolate y: y = 5x - 10 - 1, which simplifies to y = 5x - 11.
7. To find the equation of a line perpendicular to y = -3x + 7 and passing through (3, 5), we know that perpendicular lines have negative reciprocal slopes. So, the slope of the new line will be 1/3 (reciprocal of -3). Using the point-slope form again, the equation will be y - y1 = m(x - x1).
Using the values (3, 5) and the slope 1/3, the equation becomes: y - 5 = (1/3)(x - 3). Simplifying, we get y - 5 = 1/3x - 1.
To write it in slope-intercept form, we can isolate y: y = 1/3x - 1 + 5, which simplifies to y = 1/3x + 4.
8. To sketch the graph of y = |x - 3| - 2, we can start by understanding the transformations applied to the absolute value function y = |x|.
The "x - 3" inside the absolute value sign represents a horizontal shift of the graph 3 units to the right. The "- 2" outside the absolute value represents a vertical shift of the graph 2 units downward.
The graph of y = |x| is a V-shaped graph centered at the origin (0, 0). After applying the transformations, the graph of y = |x - 3| - 2 will have its vertex at (3, -2), with the V-shaped graph opening upwards. It will shift horizontally 3 units to the right and vertically down 2 units from the origin.
To sketch the graph, plot the vertex (3, -2), and then plot additional points on either side (e.g., x = 0, 1, 2, 4, 5, etc.) and find the corresponding y-values using the equation y = |x - 3| - 2. Connect the plotted points to complete the graph.
do you know what the standard form is? First step should probably be to clear the fractions: 3y = -2x - 3
Now just rearrange the terms as needed
y = 5x – 2 has slope = 5
So, now you have a slope and a point, so start with the point-slope form:
y+1 = 5(x-2)
Now rearrange as needed.
y = –3x+7 has slope -3
So, you want a line with slope 1/3
Now follow the steps of #6
y = |x| is a v-shape, with vertex at (0,0)
y = |x – 3| – 2 is the same graph, shifted right 3 and down 2.