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2 years ago

Determine whether the relation is a function. quick EMERGENCY I NEED HELP

Mathematics

1 answer:

saul85 [17]2 years ago

6 0

Not a function.
It goes from -1=-4 to -1=-2

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Find the y intercepts of 3x^2+24x-51. quadratic formula

julia-pushkina [17]

Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , with a = x^2 coefficient, b = x coefficient, and c = constant.

Firstly, starting with the y-intercept. To find the y-intercept, set the x variable to zero and solve as such:

$y=3*0^2+24*0-51\\y=0+0-51\\y=-51$

Your y-intercept is (0,-51).

Next, using our equation plug the appropriate values into the quadratic formula:

$x=\frac{-24\pm \sqrt{24^2-4*3*(-54)}}{2*3}$

Next, solve the multiplications and exponent:

$x=\frac{-24\pm \sqrt{576-(-648)}}{6}\\\\x=\frac{-24\pm \sqrt{576+648}}{6}$

Next, solve the addition:

$x=\frac{-24\pm \sqrt{1224}}{6}$

Now, simplify the radical using the product rule of radicals as such:

Product Rule of Radicals: √ab = √a × √b

√1224 = √12 × √102 = √2 × √6 × √6 × √17 = 6 × √2 × √17 = 6√34

$x=\frac{-24\pm 6\sqrt{34}}{6}$

Next, divide:

$x=-4\pm \sqrt{34}$

The exact values of your x-intercepts are (-4 + √34, 0) and (-4 - √34, 0).

Now to find the approximate values, solve this twice: once with the + symbol and once with the - symbol:

$x=-4+ \sqrt{34},-4- \sqrt{34}\\x\approx 1.83, -9.83$

The approximate values of your x-intercepts (rounded to the hundredths) are (1.83,0) and (-9.83,0).

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3 years ago

Determine the intervals on which the function is increasing, decreasing and constant

viva [34]

Answer:

increasing: (-∞, 0)
decreasing: (0, ∞)

Step-by-step explanation:

The function goes up to the right until it gets to the vertex at x=0. Then it goes down to the right. That is, it is ...

increasing from -∞ to 0 (not including 0)

decreasing from 0 to +∞ (not including 0)

_____

At x=0, the function is neither increasing nor decreasing, so x=0 is not part of either interval.

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3 years ago

Determine whether the relation is a function. quick EMERGENCY I NEED HELP ​

Determine whether the relation is a function. quick EMERGENCY I NEED HELP