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Strike441 [17]
3 years ago
9

Help me please , thank you so much

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
6 0

Domain: All real numbers, or (-∞,∞) depending on how you want to write it.

Range: [-4, ∞)

Vertex: (-2, -4)

Axis of symmetry: x = -2

Interval of increase: (-2, ∞)

Interval of decrease: (-∞, -2)

Y-int: (0, 0)

X-int/Zero/Solution(s): (-4, 0) and (0, 0)

For a polynomial function like this, which is a quadratic function, the domain is always negative infinity to positive infinity, or all real numbers.

The range is the lowest y-value to the highest y-value. The lowest y-value for this is at the vertex, but the function has no upper limit for the y-value.

The vertex is the point that you can very clearly see is at the tip of the graph.

Axis of symmetry is the vertical line that splits the graph in half at the vertex.

Interval of increase is the interval x values for which the function is increasing. The function increases to the right of the vertex.

Interval of decrease is to the left of the vertex.

The y-intercept is clearly (0, 0), where the graph intersects the y-axis.

The x-intercepts are also clear, since you can see where the graph intersects the x-axis.

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What form is this equation written in? y-7=2(x-6)
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Choose the correct simplification of (4x − 3)(3x2 − 4x − 3).
tester [92]
(4x - 3)(3x^2 - 4x - 3) 

=4x\cdot \:3x^2+4x\left(-4x\right)+4x\left(-3\right)-3\cdot \:3x^2-3\left(-4x\right)-3\left(-3\right) 

=4x\cdot \:3x^2-4x\cdot \:4x-4x\cdot \:3-3\cdot \:3x^2+3\cdot \:4x+3\cdot \:3 

\mathrm{Simplify}\:4x\cdot \:3x^2-4x\cdot \:4x-4x\cdot \:3-3\cdot \:3x^2+3\cdot \:4x+3\cdot \:3:\quad 12x^3-25x^2+9 

=12x^3-25x^2+9
8 0
3 years ago
Ex 2.8<br> 3. find the maximum value of y for the curve y=x^5 -3 for -2≤x≤1
harkovskaia [24]
y=x^5-3\\ y'=5x^4\\\\ 5x^4=0\\ x=0\\ 0\in [-2,1]\\\\ y''=20x^3\\\\&#10;y''(0)=20\cdot0^3=0

The value of the second derivative for x=0 is neither positive nor negative, so you can't tell whether this point is a minimum or a maximum. You need to check the values of the first derivative around the point.
But the value of 5x^4 is always positive for x\in\mathbb{R}\setminus \{0\}. That means at x=0 there's neither minimum nor maximum.
The maximum must be then at either of the endpoints of the interval [-2,1].
The function y is increasing in its entire domain, so the maximum value is at the right endpoint of the interval.

y_{max}=y(1)=1^5-3=-2
4 0
3 years ago
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