Answer:
yes
Step-by-step explanation:
The line intersects each parabola in one point, so is tangent to both.
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For the first parabola, the point of intersection is ...
y^2 = 4(-y-1)
y^2 +4y +4 = 0
(y+2)^2 = 0
y = -2 . . . . . . . . one solution only
x = -(-2)-1 = 1
The point of intersection is (1, -2).
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For the second parabola, the equation is the same, but with x and y interchanged:
x^2 = 4(-x-1)
(x +2)^2 = 0
x = -2, y = 1 . . . . . one point of intersection only
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If the line is not parallel to the axis of symmetry, it is tangent if there is only one point of intersection. Here the line x+y+1=0 is tangent to both y^2=4x and x^2=4y.
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Another way to consider this is to look at the two parabolas as mirror images of each other across the line y=x. The given line is perpendicular to that line of reflection, so if it is tangent to one parabola, it is tangent to both.
The answer would be D, 11 units! :)
4x - 3 = 9
4x = 9 + 3
4x = 12
x = 3
hope that helps, God bless!
Answer:
g(x) is a quadratic function ⇒ 2
Step-by-step explanation:
- The quadratic function is the function that has 2 as the greatest power of the variable
- The form of the quadratic function is f(x) = ax² + bx + c, where a, b, and c are constant
Let us use the information above to solve the question
∵ f(x) = 
∵ x is the exponent of the base 1.5
→ That means f(x) is not in the form of the quadratic function
∴ f(x) is not in the form of the quadratic function above
∴ f(x) does not represent a quadratic function
∴ f(x) is not a quadratic function
∵ g(x) = 500x² + 345x
∴ The greatest power of x is 2
→ That means g(x) is in the form of the quadratic function above
∵ g(x) is in the form of the quadratic function above, where a = 500,
b = 345, and c = 0 (constant values)
∴ g(x) represents a quadratic function
∴ g(x) is a quadratic function
Answer:
70
Step-by-step explanation:
