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1 year ago

Graphs of Function,

Mathematics

1 answer:

andrew-mc [135]1 year ago

5 0

Answer:

A function is increasing when the gradient is positive

A function is decreasing when the gradient is negative

Question 7

If you draw a tangent to the curve in the interval x < -2 then the tangent will have a positive gradient, and so the function is increasing in this interval.

If you draw a tangent to the curve in the interval x > -2 then the tangent will have a negative gradient, and so the function is decreasing in this interval.

If you draw a tangent to the curve at the vertex of the parabola, it will be a horizontal line, and so the gradient at x = -2 will be zero.

The function is increasing when x < -2

$(- \infty,-2)$

The function is decreasing when x > -2

$(-2, \infty)$

Additional information

We can actually determine the intervals where the function is increasing and decreasing by differentiating the function.

The equation of this graph is:

$f(x)=-2x^2-8x-8$

$\implies f'(x)=-4x-8$

The function is increasing when $f'(x) > 0$

$\implies -4x-8 > 0$

$\implies -4x > 8$

$\implies x < -2$

The function is decreasing when $f'(x) < 0$

$\implies -4x-8 < 0$

$\implies -4x < 8$

$\implies x > -2$

This concurs with the observations made from the graph.

Question 8

This is a straight line graph. The gradient is negative, so:

The function is decreasing for all real values of x

$(- \infty,+ \infty)$

But if they want the interval for the grid only, it would be -4 ≤ x ≤ 1

$[-4,1]$

Question 9

If you draw a tangent to the curve in the interval x < -1 then the tangent will have a negative gradient, and so the function is decreasing in this interval.

If you draw a tangent to the curve in the interval x > -1 then the tangent will have a positive gradient, and so the function is increasing in this interval.

If you draw a tangent to the curve at the vertex of the parabola, it will be a horizontal line, and so the gradient at x = -1 will be zero.

The function is decreasing when x < -1

$(- \infty,-1)$

The function is increasing when x > -1

$(-1, \infty)$

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3 years ago

For what values of the variable, do the following fractions exist: y^2-1/y+y/y-3

ale4655 [162]

Answer:

Remember that the division by zero is not defined, this is the criteria that we will use in this case.

1) $\frac{y^2 - 1}{y} + \frac{y}{y - 3}$

So the fractions are defined such that the denominator is never zero.

For the first fraction, the denominator is zero when y = 0

and for the second fraction, the denominator is zero when y = 3

Then the fractions exist for all real values except for y = 0 or y = 3

we can write this as:

R / {0} U { 3}

(the set of all real numbers except the elements 0 and 3)

2) $\frac{b + 4}{b^2 + 7}$

Let's see the values of b such that the denominator is zero:

b^2 + 7 = 0

b^2 = -7

b = √-7

This is a complex value, assuming that b can only be a real number, there is no value of b such that the denominator is zero, then the fraction is defined for every real number.

The allowed values are R, the set of all real numbers.

3) $\frac{a}{a*(a - 1) - 1}$