How to Find X and Y Intercepts of Graphs
How do you find the x- and y-intercepts of graphs of functions and equations?
The x- and y-intercepts of a graph are the points of intersection with the x-axis and the y-axis, respectively. This tutorial provides detailed, step-by-step examples showing how to calculate these points algebraically and interpret them graphically.
Example 1
Find the x- and y-intercepts of the graph of the linear function defined by:
\[ f(x) = -3x + 9 \]
Solution to Example 1
- A point on the x-axis has a y-coordinate equal to \(0\). To find the x-intercept, set \(y = f(x) = 0\) and solve for \(x\):
\[ -3x + 9 = 0 \]
- Solve for \(x\):
\[ x = 3 \]
- The x-intercept of the graph of \(f\) is:
x-intercept: \[ (3, 0) \]
- A point on the y-axis has an x-coordinate equal to zero. To find the y-intercept, evaluate \(f(0)\):
\[ f(0) = -3(0) + 9 = 9 \]
y-intercept: \[ (0, 9) \]
Example 2
Find the x- and y-intercepts of the circle given by the equation:
\[ (x - 1)^2 + (y - 2)^2 = 16 \]
Solution to Example 2
- To find the y-intercept: Set \(x = 0\) in the given equation:
\[ (0 - 1)^2 + (y - 2)^2 = 16 \implies 1 + (y - 2)^2 = 16 \]
\[ (y - 2)^2 = 15 \]
The solutions are: \[ y_1 = 2 + \sqrt{15} \quad \text{and} \quad y_2 = 2 - \sqrt{15} \]
- To find the x-intercept: Set \(y = 0\) in the given equation:
\[ (x - 1)^2 + (0 - 2)^2 = 16 \implies (x - 1)^2 + 4 = 16 \]
\[ (x - 1)^2 = 12 \]
The solutions are: \[ x_1 = 1 - \sqrt{12} \quad \text{and} \quad x_2 = 1 + \sqrt{12} \]
- The x- and y-intercepts of the circle are:
x-intercepts: \[ A = (1 - \sqrt{12}, 0) \quad \text{and} \quad B = (1 + \sqrt{12}, 0) \]
y-intercepts: \[ C = (0, 2 - \sqrt{15}) \quad \text{and} \quad D = (0, 2 + \sqrt{15}) \]
- The graph shown below illustrates the circle and its intercepts. Notice how graphical readings match the calculated coordinates.
Example 3
Calculate the x- and y-intercepts of the linear equation:
\[ 3x + 2y = 6 \]
Solution to Example 3
- x-intercept: Set \(y = 0\):
\[ 3x + 2(0) = 6 \implies 3x = 6 \implies x = 2 \]
x-intercept: \[ A = (2, 0) \]
- y-intercept: Set \(x = 0\):
\[ 3(0) + 2y = 6 \implies 2y = 6 \implies y = 3 \]
y-intercept: \[ B = (0, 3) \]
- The graph of the equation is shown below, confirming the calculated intercepts:
Example 4
Calculate the x- and y-intercepts of the quadratic function:
\[ f(x) = -x^2 + 2x + 3 \]
Solution to Example 4
- y-intercept: Set \(x = 0\):
\[ y = f(0) = -0^2 + 2(0) + 3 = 3 \]
- x-intercepts: Set \(y = f(x) = 0\):
\[ -x^2 + 2x + 3 = 0 \implies x^2 - 2x - 3 = 0 \]
Factor the quadratic:
\[ (x - 3)(x + 1) = 0 \implies x_1 = -1 \text{ and } x_2 = 3 \]
- Summary:
x-intercepts: \(A = (-1, 0)\), \(B = (3, 0)\)
y-intercept: \(C = (0, 3)\)
- The graph of the function and its intercepts are shown below:
Example 5
Determine the x- and y-intercepts of the logarithmic function:
\[ f(x) = -\ln(x + 1) - 2 \]
Solution to Example 5
- y-intercept: Set \(x = 0\):
\[ y = f(0) = -\ln(0 + 1) - 2 = -2 \implies B = (0, -2) \]
- x-intercept: Set \(y = f(x) = 0\):
\[ -\ln(x + 1) - 2 = 0 \implies \ln(x + 1) = -2 \]
Convert to exponential form:
\[ x + 1 = e^{-2} \implies x = -1 + \frac{1}{e^2} \]
x-intercept: \[ A = \left(-1 + \frac{1}{e^2}, 0\right) \]
- The graph of the function is shown below:
Example 6
Calculate the x- and y-intercepts of the exponential function:
\[ f(x) = e^{x + 1} - 2 \]
Solution to Example 6
- y-intercept: Evaluate \(f(0)\):
\[ y = f(0) = e^{0 + 1} - 2 = e - 2 \implies B = (0, e - 2) \]
- x-intercept: Set \(f(x) = 0\):
\[ e^{x + 1} - 2 = 0 \implies e^{x + 1} = 2 \]
Take the natural logarithm of both sides:
\[ x + 1 = \ln 2 \implies x = -1 + \ln 2 \]
x-intercept: \[ A = (-1 + \ln 2, 0) \]
- The graph of the function and its intercepts are shown below:
Example 7
Calculate the x- and y-intercepts of the rational function:
\[ f(x) = \frac{x^2 - x - 2}{x^2 - x - 3} \]
Solution to Example 7
- y-intercept: Evaluate \(f(0)\):
\[ y = f(0) = \frac{0^2 - 0 - 2}{0^2 - 0 - 3} = \frac{-2}{-3} = \frac{2}{3} \implies C = \left(0, \frac{2}{3}\right) \]
- x-intercepts: Set the numerator equal to zero:
\[ x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0 \]
\[ x_1 = -1 \quad \text{and} \quad x_2 = 2 \]
- Summary:
x-intercepts: \(A = (-1, 0)\), \(B = (2, 0)\)
y-intercept: \(C = \left(0, \frac{2}{3}\right)\)
- The graph of the rational function is shown below:
Example 8
Calculate the x- and y-intercepts of the sine function:
\[ f(x) = \sin(x) + \frac{1}{2} \]
Solution to Example 8
- y-intercept: Evaluate \(f(0)\):
\[ y = f(0) = \sin(0) + \frac{1}{2} = \frac{1}{2} \implies D = \left(0, \frac{1}{2}\right) \]
- x-intercepts: Set \(f(x) = 0\):
\[ \sin(x) + \frac{1}{2} = 0 \implies \sin(x) = -\frac{1}{2} \]
Due to periodicity, the general solutions are:
\[ x_1 = \frac{7\pi}{6} + 2k\pi, \quad k = 0, \pm1, \pm2, \ldots \]
\[ x_2 = \frac{11\pi}{6} + 2k\pi, \quad k = 0, \pm1, \pm2, \ldots \]
- Key coordinate points shown on the graph include:
x-intercepts: \(A = \left(-\frac{\pi}{6}, 0\right)\), \(B = \left(\frac{7\pi}{6}, 0\right)\), \(C = \left(\frac{11\pi}{6}, 0\right)\)
y-intercept: \(D = \left(0, \frac{1}{2}\right)\)
- The graph of the periodic wave function and its intercepts are shown below:
