Make an accuwe graph of the piecewise-defined function For the quadratic function f(x) = x + 2x – 8: a) (5 pts.) Determine the coordinates of the vertex algebraically b) (4 pts. Use the result from a) to determine the intervals where the function is increasing and where it is decreasing. Use interval notation. (5 pts. Find the zeroes of the function, and determine the x-intercepts. x²+27-8 O=x+28 (x+2)(x-4) In (2016-410) Cup. d) (1 pt.) Determine the y-intercepts (4 pts.) Sketch the graph of the function and label the special points. 0,8 (3 pts.) Use the difference quotient to determine the slope of the secant line, when x changes from x = 1 to x = 5.
The Correct Answer and Explanation is :
It looks like you have a piecewise-defined function, but the specific function isn’t entirely clear from the way the problem is written. I’ll assume you’re working with a quadratic function ( f(x) = x^2 + 2x – 8 ), based on the expression and the other information in your question. Let’s go step-by-step to answer your questions.
a) Determining the coordinates of the vertex algebraically
The quadratic function is given by ( f(x) = x^2 + 2x – 8 ). To find the vertex of a parabola, we can use the formula for the x-coordinate of the vertex:
[
x_{\text{vertex}} = \frac{-b}{2a}
]
For the function ( f(x) = ax^2 + bx + c ), we have ( a = 1 ), ( b = 2 ), and ( c = -8 ). Plugging in these values:
[
x_{\text{vertex}} = \frac{-2}{2(1)} = -1
]
Now, substitute ( x = -1 ) back into the function to find the y-coordinate of the vertex:
[
f(-1) = (-1)^2 + 2(-1) – 8 = 1 – 2 – 8 = -9
]
So, the coordinates of the vertex are ( (-1, -9) ).
b) Intervals where the function is increasing or decreasing
Since the parabola opens upwards (because ( a = 1 > 0 )), the function is:
- Decreasing on the interval ( (-\infty, -1) ), where ( x ) is less than the x-coordinate of the vertex.
- Increasing on the interval ( (-1, \infty) ), where ( x ) is greater than the x-coordinate of the vertex.
c) Finding the zeroes of the function (x-intercepts)
To find the zeroes, we set ( f(x) = 0 ) and solve for ( x ):
[
x^2 + 2x – 8 = 0
]
We can solve this quadratic equation using factoring:
[
(x + 4)(x – 2) = 0
]
Setting each factor equal to zero gives:
[
x + 4 = 0 \quad \Rightarrow \quad x = -4
]
[
x – 2 = 0 \quad \Rightarrow \quad x = 2
]
So, the x-intercepts (zeroes) are at ( x = -4 ) and ( x = 2 ).
d) Determining the y-intercept
The y-intercept occurs when ( x = 0 ). Substitute ( x = 0 ) into the function:
[
f(0) = (0)^2 + 2(0) – 8 = -8
]
So, the y-intercept is at ( (0, -8) ).
e) Sketching the graph
To sketch the graph, we plot the following points:
- Vertex: ( (-1, -9) )
- x-intercepts: ( (-4, 0) ) and ( (2, 0) )
- y-intercept: ( (0, -8) )
Since the parabola opens upwards and has the vertex at ( (-1, -9) ), the graph will be symmetric around the vertical line ( x = -1 ).
f) Using the difference quotient to find the slope of the secant line from ( x = 1 ) to ( x = 5 )
The difference quotient is given by:
[
\frac{f(x_2) – f(x_1)}{x_2 – x_1}
]
where ( x_1 = 1 ) and ( x_2 = 5 ). We need to evaluate ( f(1) ) and ( f(5) ):
[
f(1) = (1)^2 + 2(1) – 8 = 1 + 2 – 8 = -5
]
[
f(5) = (5)^2 + 2(5) – 8 = 25 + 10 – 8 = 27
]
Now, calculate the difference quotient:
[
\frac{f(5) – f(1)}{5 – 1} = \frac{27 – (-5)}{5 – 1} = \frac{32}{4} = 8
]
So, the slope of the secant line is 8.
Conclusion
To summarize:
- The vertex is ( (-1, -9) ).
- The function is decreasing on ( (-\infty, -1) ) and increasing on ( (-1, \infty) ).
- The x-intercepts are ( x = -4 ) and ( x = 2 ).
- The y-intercept is ( (0, -8) ).
- The slope of the secant line between ( x = 1 ) and ( x = 5 ) is 8.
This should help you understand the behavior of the quadratic function and its graph!