Right
for (b) I would have written ( -1 , 3 )
use the point-slope form linear equation y-3=3(x+1)
a. slope is 3
b. point this line passes through that is basis of this equatio y = 3 x = -1
c. rewrite this equation in slope-intercept form
y = mx+b
y-3=3(x+1)
y-3+3 = 3x +3 +3
y = 3x + 6
d. what is y intercept of line? 6
thanks for helping
for (b) I would have written ( -1 , 3 )
b. To find the point that the line passes through, you need to look at the equation in point-slope form, which is y - y1 = m(x - x1). In this case, the equation y - 3 = 3(x + 1) has a slope (m) of 3 and the point (-1, 3) [where x = -1 and y = 3] is given. Therefore, the line passes through the point (-1, 3).
c. To rewrite the equation in slope-intercept form (y = mx + b), you need to isolate the y-term. Start with the given equation y - 3 = 3(x + 1). Distribute 3 to both terms inside the parentheses: y - 3 = 3x + 3. Then, add 3 to both sides of the equation: y - 3 + 3 = 3x + 3 + 3. Simplify: y = 3x + 6. So, the equation in slope-intercept form is y = 3x + 6.
d. The y-intercept of the line is the value of y when x is zero. From the slope-intercept form equation, y = 3x + 6, it can be observed that the y-intercept is 6. Therefore, the y-intercept of the line is 6.