Without sketching the graph, find the x-intercepts and y-intercepts of the graph of the equation.
[tex]y=x^2-2 x-15[/tex]
(On this one, just give me either one of the two XINTERCEPTS.)
Answer (1)
Set y = 0 in the equation y = x 2 − 2 x − 15 .
Factor the quadratic equation x 2 − 2 x − 15 = 0 to get ( x − 5 ) ( x + 3 ) = 0 .
Solve for x to find the x-intercepts: x = 5 or x = − 3 .
State one of the x-intercepts: 5 .
Explanation
Finding the x-intercepts To find the x-intercepts of the graph of the equation y = x 2 − 2 x − 15 , we need to find the values of x when y = 0 . This means we need to solve the quadratic equation x 2 − 2 x − 15 = 0 .
Factoring the quadratic equation We can solve the quadratic equation by factoring. We are looking for two numbers that multiply to -15 and add to -2. These numbers are -5 and 3. So, we can factor the equation as ( x − 5 ) ( x + 3 ) = 0 .
Solving for x Now, we set each factor equal to zero and solve for x :
x − 5 = 0 or x + 3 = 0
Solving these equations gives us:
x = 5 or x = − 3
So, the x-intercepts are 5 and -3.
Final Answer We were asked to provide only one of the x-intercepts. Therefore, one of the x-intercepts is 5.
Examples
Understanding x-intercepts is crucial in various real-world applications. For instance, consider a projectile's trajectory modeled by a quadratic equation, where 'y' represents the height and 'x' represents the horizontal distance. The x-intercepts indicate where the projectile lands (when the height 'y' is zero). If the equation of the projectile's path is y = x 2 − 2 x − 15 , finding the x-intercepts helps determine the landing points. By setting y = 0 and solving for x , we find the distances at which the projectile hits the ground. This concept is vital in fields like sports (calculating ball trajectories), engineering (designing projectile launchers), and physics (analyzing motion under gravity).
