Q. Find the y-intercept of the line \(y=-\frac{15}{4}x-18\)
Answer
Set \(x=0\):
\[
y=-\frac{15}{4}\cdot 0-18=-18
\]
Thus the y-intercept is \(\boxed{(0,-18)}\).
Detailed Explanation
Problem: Find the y-intercept of the line given by the equation \( y = -\frac{15}{4}x – 18 \).
Step 1 – Recall the definition of the y-intercept
The y-intercept is the point where the line crosses the y-axis. Every point on the y-axis has \( x = 0 \). To find the y-intercept substitute \( x = 0 \) into the equation and solve for \( y \).
Step 2 – Substitute \( x = 0 \)
\[
y = -\frac{15}{4}\cdot 0 – 18
\]
Step 3 – Evaluate
\[
-\frac{15}{4}\cdot 0 = 0
\]
\[
y = 0 – 18 = -18
\]
Answer
\[
\text{The y-intercept point is }(0,-18)\quad\text{and the y-intercept value is }-18.
\]
See full solution
FAQs
What is the y-intercept of the line y = -tfrac{15}{4}x - 18?
Set x = 0: y = -18. The y-intercept value is -18 and the intercept point is (0, -18).
How do I identify the y-intercept from slope-intercept form y = mx + b?
In y = mx + b, the constant b is the y-intercept. Here m = -tfrac{15}{4} and b = -18, so the y-intercept is -18.
What is the slope of the line y = -tfrac{15}{4}x - 18?
The slope is the coefficient of x: m = -tfrac{15}{4}, meaning the line falls 15 units for every 4 units it moves right.
How can I quickly graph the line using slope and y-intercept?
Plot the y-intercept (0, -18). From there use slope -tfrac{15}{4}: move right 4 and down 15 to get a second point (4, -33), then draw the line through them.
How do I find the x-intercept of y = -tfrac{15}{4}x - 18?
Set y = 0 and solve: 0 = -tfrac{15}{4}x - 18 → x = -18 times left(-tfrac{4}{15}right) = -tfrac{24}{5}. The x-intercept is (-tfrac{24}{5}, 0).
How do I write this line in standard form Ax + By = C?
Multiply to clear fractions: 4y = -15x - 72. Rearranged: 15x + 4y = -72 (standard form with integers).
What does a negative y-intercept tell me about the graph?
A negative y-intercept means the line crosses the y-axis below the origin. Here the crossing point is at y = -18, so the graph is well below the origin at x = 0.
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