Answer:
A) x² + 2x + 6
B) x² + 2x - 7
C) ¼(x²+2x+1))
D) 6x²+12x+6
E) -x²-2x-1
Step-by-step explanation:
A) f(x) + 5 =x²+2x+1 + 5
= x² + 2x + 6
B) f(x)-8,=x^2+2x+1-8
= x² + 2x - 7
C) ¼f(x) = ¼(x²+2x+1)
D) 6f(x) = 6(x²+2x+1) = 6x²+12x+6
E) -f(x) = -(x²+2x+1) = -x²-2x-1
I don't know where you got those answers, but the actual equation that represents those points is: y = -1/2x + 11. None of those listed above are correct.
Answer:
2
Step-by-step explanation:
Since there are four negative signs, we have -1 multiplying each other 4 times, multiplying by positive 2. This is then 1 * 2, which is 2.
Compute the slope of the line through (-2, -9) and (-4, -17):
Any line perpendicular to this line will have slope -1/4. The one that passes through (0, 10) satisifes
Answer:
In a quadratic equation of the shape:
y = a*x^2 + b*x + c
we hate that the discriminant is equal to:
D = b^2 - 4*a*c
This thing appears in the Bhaskara's formula for the roots of the quadratic equation:
You can see that the determinant is inside a square root, this means that if D is smaller than zero we will have imaginary roots (the graph never touches the x-axis)
If D = 0, the square root term dissapear, and this implies that both roots of the equation are the same, this means that the graph touches the x axis in only one point, wich coincides with the minimum/maximum of the graph)
If D > 0 we have two different roots, so the graph touches the x-axis in two different points.