Note that the 2nd equation can be re-written as y=8x-10.
According to the second equation, y=x^2+12x+30.
Equate these two equations to eliminate y:
8x-10 = x^2+12x+30
Group all terms together on the right side. To do this, add -8x+10 to both sides. Then 0 = x^2 +4x +40. You must now solve this quadratic equation for x, if possible. I found that this equation has NO REAL SOLUTIONS, so we must conclude that the given system of equations has NO REAL SOLUTIONS.
If you have a graphing calculator, please graph 8x-10 and x^2+12x+30 on the same screen. You will see two separate graphs that do NOT intersect. This is another way in which to see / conclude that there is NO REAL SOLUTION to this system of equations.
Solution:
Given:
The zeros of a function are the values of x when f(x) is equal to 0.
Hence,
Therefore, x = 1
The correct answer is OPTION A.
I believe it would be (-4, 8); because when x = -4, y = f(x) = 8.
Answer: 0.1875
Step-by-step explanation:
Change in y / change in x
(6 - 4) / (2 - 0)
2 / 2
Numerator = 2.
