+96 1. Consider the parabola y=-5x^2+17x-12 . What is the vertex of this parabola?
2. What is the equation of the line of symmetry of the parabola y=-5x^2+17x-12
3.What are the x-coordinate(s) of all point(s) where the parabola y=-5x^2+17x-12 intersects the line y=0?
4. For what values of x is -5x^2+17x-12 negative?
Sorry for all of these...
+41 1. (1.7, 2.45)
2. x = 1.7
3. (1,0) and (2.4,0)
4. (2.45, negative infinity)
+41 1. use the formula of line of symetry to find x cordanite: -b/2a
2. When you find what -b/2a is, plug it into the formula
3. set the equation to zero and find the roots
4. since the parobla has max, it is (2.45, -infinity)
+129852 1. Consider the parabola y = -5x^2+17x-12 . What is the vertex of this parabola?
We have the form Ax^2 + Bx + C
The x coordinate of the vertex is given by -B / [2A] = -17/ [ 2(-5)] = 17/10
Put this value back into the function to get the y coordinate of the vertex
-5(17/10)^2 + 17(17/10) - 12 =
-5 (289/100) + 289/10 - 12 = 49/20
So.....the vertex is ( 17/10, 49/20)
2. What is the equation of the line of symmetry of the parabola y=-5x^2+17x-12
Easy.....the equation for the symmetry line is
x = x coordinate of the vertex
So.... x = 17/10
3.What are the x-coordinate(s) of all point(s) where the parabola y=-5x^2+17x-12 intersects the line y=0?
We need to solve this
-5x^2 + 17x - 12 = 0 factor as
(-5x + 12) ( x - 1) = 0
Set each factor to 0 and solve for x and we have that
-5x + 12 = 0 x - 1 = 0
12 = 5x x = 1
12/5 = x
4. For what values of x is -5x^2+17x-12 negative?
The parabola has its vertex above the x axis and it turns downward because of the -5 in front of the x^2 term
It will intersect the x axis at x = 1 and x = 12/5
So......it will be negative from (-infinity , 1) and from (12/5, infinity)
See the graph here : https://www.desmos.com/calculator/visfr9nhfz
