SOLUTION: Could you please tell me if these are correct. Thank-you! 1. Give exact and approximate solutions to three decimal places. (x+2)^2=64 Answer: x=6,-10 2. Find the vertex f(x)=1

Algebra ->  Rational-functions -> SOLUTION: Could you please tell me if these are correct. Thank-you! 1. Give exact and approximate solutions to three decimal places. (x+2)^2=64 Answer: x=6,-10 2. Find the vertex f(x)=1      Log On


   

Question 197324: Could you please tell me if these are correct. Thank-you!
1. Give exact and approximate solutions to three decimal places.
(x+2)^2=64
Answer: x=6,-10
2. Find the vertex
f(x)=1/3(x+5)^2+6
Answer: (-10,0)
3.Find the x-and y- intercepts
f(x)=16x^2+24x+9
Answer x=3/4,y=9
4.Find and label the vertex and the line of symmetry.
f(x)=2(x-2)^2
Answer (2,0)
5. Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic function, and graph the function.
f(x)=6-x^2
Answer: 0,-x^2
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!

Problem 1 is spot on but you have errors in the rest of them.

2.



Rewrite as:



Now you have a parabola of the form:



which has a vertex at

3.

The factors are so

4.

You found the correct vertex, but you didn't give the equation for the axis of symmetry. The equation for the axis of symmetry of a parabola with a vertical axis is where is the x-coordinate of the vertex.

5.

Put it in standard form:



The vertex x-coordinate is

and the vertex y-coordinate is You need to calculate

again the equation of the axis of symmetry of a parabola with a vertical axis is where is the x-coordinate of the vertex.

The lead coefficient is < 0, so the parabola opens downward, making the value of the function at the vertex a maximum value -- hence the maximum value is the y-coordinate of the vertex.

John