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...

Pushy Jun 24, 2019

#1**+1 **

1. (1.7, 2.45)

2. x = 1.7

3. (1,0) and (2.4,0)

4. (2.45, negative infinity)

DungeyDabs Jun 24, 2019

edited by
Guest
Jun 24, 2019

#2**+1 **

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)

DungeyDabs
Jun 24, 2019

#4**+2 **

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

CPhill Jun 24, 2019