Answer:
(- 1, 1) and (- 5, - 35)
Step-by-step explanation:
since both equations express y in terms of x, equate the right sides
x² + 15x + 15 = - x² + 3x + 5 ( subtract - x² + 3x + 5 from both sides )
2x² + 12x + 10 = 0 ( divide all terms by 2 )
x² + 6x + 5 = 0 ← in standard form
(x + 1)(x + 5) = 0 ← in factored form
equate each factor to zero and solve for x
x + 1 = 0 ⇒ x = - 1
x + 5 = 0 ⇒ x = - 5
substitute these values into either of the 2 equations and solve for y
x = - 1 : y = (- 1)² + 15(- 1) + 15 = 1 - 15 + 15 = 1 ⇒ (- 1, 1 )
x = - 5 : y = (- 5)² + 15(- 5) + 15 = 25 - 75 + 15 = - 35 ⇒ (- 5, - 35 )
None of the given options
Problem
For a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height, y, and time, x , when the projectile reaches the ground
Solution
We know that the x coordinate of a quadratic function is given by:
Vx= -b/2a
And the y coordinate correspond to the maximum value of y.
Then the best options are C and D but the best option is:
D) The maximum height is a y coordinate of the vertex of the quadratic function, which occurs when x = -b/2a
The projectile reaches the ground when the height is zero. The time when this occurs is the x-intercept of the zero of the function that is farthest to the right.
Answer:
6 + 7 = 13
13 - 7 = 6
13 - 6 = 7
7 + 6 = 13
Step-by-step explanation:
Answer:It's A I think if not try B
Step-by-step explanation: