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3 years ago

Dan said that if a quadratic equation does not have any real solutions , it is not a function . Critique his reasoning

1 answer:

8 0

<span> Solutions of a quadratic functions are x-intercepts, or when y equals zero. A graph with its maximum (highest point) below the x-axis or a graph with its minimum (lowest point) above the x-axis never touches or crosses the x-axis, so it has no real solutions. It is still a quadratic function, as long as it has a variable to the second power (definition of quadratic) and passes the vertical line test (definition of function).

Hope this helped! <3

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Which of the tables represents a function? (5 points)

AlladinOne [14]

Answer:

It's table D

Step-by-step explanation:

I just took the test.

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3 years ago

How many groups of 2.4 will divide into 29.52 ?

Bas_tet [7]

29.52/2.4 =12.3
so 2.4 groups will go into 29.52, 12.3 times

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3 years ago

Find the area between the graph of the function and the x-axis over the given interval, if possible.

vodomira [7]

Answer:

$A= -\frac{5}{-1} - \lim_{x\to\infty} \frac{5}{x-2} = 5-0 = 5$

So then the integral converges and the area below the curve and the x axis would be 5.

Step-by-step explanation:

In order to calculate the area between the function and the x axis we need to solve the following integral:

$A = \int_{-\infty}^1 \frac{5}{(x-2)^2}$

For this case we can use the following substitution $u = x-2$ and we have $dx = du$

$A = \int_{a}^b \frac{5}{u^2} du = 5\int_{a}^b u^{-2}du$

And if we solve the integral we got:

$A= -\frac{5}{u} \Big|_a^b$

And we can rewrite the expression again in terms of x and we got:

$A = -\frac{5}{x-2} \Big|_{-\infty}^1$

And we can solve this using the fundamental theorem of calculus like this:

$A= -\frac{5}{-1} - \lim_{x\to\infty} \frac{5}{x-2} = 5-0 = 5$

So then the integral converges and the area below the curve and the x axis would be 5.

7 0

3 years ago

Find the equation of the axis of symmetry and the coordinates of vertex of the functions =2x^2+4x-3

SVEN [57.7K]

Hello,

Find the equation of the axis of symmetry is like find the maximum or minimum point of this function.

$f(x)=2x^2+4x-3\\\\\alpha=-\dfrac{b}{2a}=-\dfrac{4}{4}=-1\\\\$

The equation of the axis of symmetry is then : $x=-1$

$\beta=f(\alpha)=2(-1)^2+4(-1)-3=2-4-3=-5$

Coordinates of vertex : $(\alpha,\beta)=(-1;-5)$

8 0

3 years ago

30pts! Brainliest! Calculus 2!

masha68 [24]

Substitute 1/x with u
dx = -x^2 du
-1/4 integral of e^u•du
Apply exponential rule -(e^u)/4
Undo the substitution u is 1/x
Answer is (-(e^1/x)/4 ) + C

4 0

3 years ago