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Arada [10]
2 years ago
5

Graphs of Function,

Mathematics
1 answer:
andrew-mc [135]2 years ago
5 0

Answer:

A function is increasing when the gradient is positive

A function is decreasing when the gradient is negative

<u>Question 7</u>

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)

<u>Additional information</u>

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.

<u>Question 8</u>

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]

<u>Question 9</u>

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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Which of the equations below could be the equation of this parabola?
nirvana33 [79]

Answer:

 y=-4x^2  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

y=-4x^2

\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)

As

\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)

\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}

y=-4x^2

\mathrm{The\:parabola\:params\:are:}

a=-4,\:m=0,\:n=0

x_v=\frac{m+n}{2}

x_v=\frac{0+0}{2}

x_v=0

\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-4\cdot \:0^2

y_v=0

Therefore, the parabola vertex is

\left(0,\:0\right)

\mathrm{If}\:a

\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

a=-4

\mathrm{Maximum}\space\left(0,\:0\right)

so,

\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)

Therefore,  y=-4x^2  is the equation of this parabola. The graph is also attached.

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