Answer:
y- intercept --> Location on graph where input is zero
f(x) < 0 --> Intervals of the domain where the graph is below the x-axis
x- intercept --> Location on graph where output is zero
f(x) > 0 --> Intervals of the domain where the graph is above the x-axis
Step-by-step explanation:
Y-intercept: The y-intercept is equivalent to the point where x= 0. 'x' is the input variable in an equation, therefore the y-intercept is where the input, or x, is equal to 0.
f(x) <0: Notice the 'lesser than' sign. This means that the value of f(x), or 'y', is less than 0. This means that this area consists of intervals of the domain below the x-axis.
X-intercept: The x-intercept is the location of the graph where y= 0, or the output is equal to 0.
f(x) >0: In this, there is a 'greater than' sign. This means that f(x), or 'y', is greater than 0. Therefore, this consists of intervals of the domain above the x-axis.
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.
Plug x and y when you type in the calculator
2-2(2+3) and see your answer
Where the points intersect should be the solution to systems of equations.
Answer:
100 Rs 5 coins and 50 Rs 2 coins
Step-by-step explanation:
Bhairav collected Rs.600 in his piggy bank by putting in Rs.2 and Rs.5 coins. The number of Rs.5 coins are twice as many as Rs.2 coins
Let x be the number of 5 coins
and y be the number of 2 coins
The number of Rs.5 coins are twice as many as Rs.2 coins


Replace x with 2y


So 100 Rs 5 coins and 50 Rs 2 coins