urgent math
posted by Anne
without graphing, find the vertex, the azis of symemtry, and the maximum value or the minium value.
Respond to this Question
Similar Questions

math
A parabola has the equation y = 4(x3)^27 Choose 2 true statements: A) The parabola has a minimum value B) The parabola has a maximum value C) The parabola does not cross the yaxis D) The parabola does not cross the xaxis E) The … 
math
A parabola has the equation y = 4(x3)^27 Choose 2 true statements: A) The parabola has a minimum value B) The parabola has a maximum value C) The parabola does not cross the yaxis D) The parabola does not cross the xaxis E) The … 
Algebra2
Complete parts a – c for each quadratic function: a. Find the yintercept, the equation of the axis of symmetry and the x coordinate of the vertex. b. Make a table of values that includes the vertex. c. Use this information to graph … 
math
find the vertex, the line of symmetry and the maximum or minimum value of f(x). graph the function. f(x)=(x+6)^22 The vertex is The line of symmetry is x= The minimum/maximum value of f(x) is? 
Math/Algebra...can you check this please...
f(x)= 4(x+5)^2+3 The vertex is : 5,3 The line of symmetry is x= 3 The maximum/minimum value of f(x)= 5 Is the value of f(5)=3, a minimum or maximum? 
College algebra
Determine without graphing, whether the given function has a maximum value and then find the value. f(x)=3x^2+18x6 
Determine without graphing
Determine without graphing, whether the given function has a maximum value and then find the value. f(x)=3x^2+30x1 
Math
How to find coordinates of vertex, and find out if the function has minimum or maximum value and find that value? 
Algebra Help
Determine without graphing, whether the given quadratic function has a max value or min value and then find the value. F(x)=2x^2+16x1 
Calculus
Find the minimum value of the following function: h(x)= x(12(x+1)^(1/3) 12) on the interval [0, 26] Honestly, I have found by graphing that the minimum value is at the coordinate point (9,3), but I need to do this in a way that …