Which graph could represent the function [tex]f(x)=(x+3.8)^2-2.7[/tex]?
Answer (1)
The given function is f ( x ) = ( x + 3.8 ) 2 − 2.7 , which is a quadratic function in vertex form.
Identify the vertex as ( − 3.8 , − 2.7 ) and note that the parabola opens upwards since the coefficient of the ( x + 3.8 ) 2 term is positive.
Calculate the y-intercept by setting x = 0 , which gives f ( 0 ) = ( 3.8 ) 2 − 2.7 = 11.74 .
The graph should have a vertex at ( − 3.8 , − 2.7 ) , open upwards, and have a y-intercept at ( 0 , 11.74 ) . Therefore, the final answer is: The graph with vertex ( − 3.8 , − 2.7 ) opening upwards and y-intercept at ( 0 , 11.74 )
Explanation
Analyze the quadratic function We are given the quadratic function f ( x ) = ( x + 3.8 ) 2 − 2.7 and we need to determine which graph represents this function. The function is in vertex form, which makes it easy to identify the vertex of the parabola. The vertex form of a quadratic function is given by f ( x ) = a ( x − h ) 2 + k , where ( h , k ) is the vertex of the parabola. In our case, a = 1 , h = − 3.8 , and k = − 2.7 . Thus, the vertex is ( − 3.8 , − 2.7 ) . Since 0"> a = 1 > 0 , the parabola opens upwards. This means that the vertex is the minimum point of the parabola.
Find the y-intercept To further analyze the graph, we can find the y-intercept by setting x = 0 in the equation f ( x ) = ( x + 3.8 ) 2 − 2.7 . This gives us:
f ( 0 ) = ( 0 + 3.8 ) 2 − 2.7 = ( 3.8 ) 2 − 2.7
Calculating this value:
f ( 0 ) = 14.44 − 2.7 = 11.74
So the y-intercept is ( 0 , 11.74 ) .
Determine the graph Now we know the vertex of the parabola is ( − 3.8 , − 2.7 ) , the parabola opens upwards, and the y-intercept is ( 0 , 11.74 ) . We need to select the graph that matches these characteristics. The vertex is in the third quadrant, and the y-intercept is positive. The parabola opens upwards from the vertex.
Examples
Quadratic functions are used in various real-world applications, such as modeling the trajectory of a projectile, designing parabolic mirrors and reflectors, and optimizing shapes in engineering. For example, when throwing a ball, the path it follows can be closely approximated by a parabola. Understanding the vertex and intercepts of a quadratic function helps predict the maximum height the ball will reach and where it will land.
