1) slope is 6 and y-intercept is ( 0,5) y = mx + b, m = 6, b = 5 y = 6x + 5 2)line passes through the points ( 3,6) and ( 6,3 ) First find the slope: m = (3-6)/(6-3) = -3/3 = -1 y = -x + b Plug in one of the given points (x,y) and find b 6 = -3 + b 9 = b y = -x + 9 a horizontal line that passes through the point ( -1,7)Horizontal lines have a constant y-value and formaty = c where c is a constant number. y = 7 y=-3x+3x intercept: set y = 0 and solve for x0 = -3x + 33x = 3x = 1x-intercept: (1, 0) y-intercept: set x = 0 and solve for yy = -3(0) + 3y = 3y-intercept: (0,3) y=0,5x-1Is this two equations? The line y=0 has y-intercept at (0,0)The x-intercept is the entire x-axis y=5x-1x -intercept: Set y = 0 and solve for x y-intercept: Set x = 0 and solve for y

5 0

2 years ago

F(x) = square root 2x

dezoksy [38]

$\\ \sf\longmapsto (f.g)(x)$

$\\ \sf\longmapsto f(x).g(x)$

$\\ \sf\longmapsto \sqrt{2x}.\sqrt{50x}$

$\\ \sf\longmapsto \sqrt{100x^2}$

$\\ \sf\longmapsto 10x$

5 0

1 year ago

The graphs below have the same shape. What is the equation of the red

anyanavicka [17]

Answer:

Option B is correct .

Step-by-step explanation:

According to Question , both the graph have same shape . If we look at the the first graph it cuts x - axis at (0 , 2) and ( 0 , -2) . Hence x = 2 and -2 are the zeroes of the equation .

And ,the given function is ,

$\implies f(x) = 4 - x^2 \\\\\implies f(x) = 4-x^2=0 \\\\\implies 2^2-x^2=0\\\\\implies (2-x)(2+x) = 0 \\\\\boxed{\red{\bf \implies x = 2 , (-2) }}$

Hence ,we can can see that x = 2 and (-2) are the zeroes of graph. 

This implies that if we know the zeroes , we can frame the Equation.

On looking at second parabola , it's clear that cuts x - axis at ( 1, 0 ) and (-1,0). So , 1 and -1 are the zeroes of the quadratic equation . Let the function be g(x) . Here , a and ß are the zeroes.

$\implies g(x) = (x-\alpha)(x-\beta) \\\\\implies g(x) = (1+x)(1-x) \\\\\boxed{\pink{\bf \implies g(x) = 1 - x^2}}$

Hence option B is correct .

8 0

2 years ago

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