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3 years ago

A graph of a quadratic function is shown below.

Mathematics

1 answer:

stira [4]3 years ago

6 0

Answer:

d. (10,800)

Step-by-step explanation:

Vertex appears to be at x = 10

Vertex: (10,800)

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Help me with this please!!

ser-zykov [4K]

Answer:

-1.3

Step-by-step explanation:

The graph shows the straight line passing through the points (3,-1) and (-3,7).

The slope of the line passing through the points $(x_1,y_1)$ and $(x_2,y_2)$ is

$\dfrac{y_2-y_1}{x_2-x_1}$

In your case, the slope is

$\dfrac{7-(-1)}{-3-3}\\ \\=\dfrac{7+1}{-6}\\ \\=-\dfrac{8}{6}\\ \\=-\dfrac{4}{3}\\ \\\approx -1.3$

The rate of change is -1.3

8 0

3 years ago

19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical

Tems11 [23]

EXPLANATION

Since we have the function:

$f(x)=\frac{x^2-8x+15}{x^2}$

Vertical asymptotes:

$For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.$

Taking the denominator and comparing to zero:

$x+5=0$

The following points are undefined:

$x=-5$

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

$\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.$ $If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.$ $If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}$ $\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}$ $\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1$ $\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}$ $\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1$ $y=\frac{1}{1}$ $\mathrm{The\:horizontal\:asymptote\:is:}$ $y=1$

In conclusion: