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nata0808 [166]

3 years ago

Mathematics

2 answers:

dedylja [7]3 years ago

7 0

Answer with explanation:

⇒Domain:

f(x)= -x² -2 x +15

= - (x²+2 x -15)

Splitting the middle term

= - (x²+5 x - 3 x -15)

= -[ x × (x+5) -3× (x+5)]

= -(x-3)(x+5)

y=f(x)=(3 -x)(x+5)

Domain of the function is defined as set of all values of x, for which y is defined.

f(x) is defined as all real values of x.So, Domain = R.

⇒Range:

$y=-x^2-2 x +15\\\\y=-(x^2+2 x-15)\\\\ y=-[(x+1)^2-1-15]\\\\y= -(x+1)^2+16\\\\ 16 -y=(x+1)^2\\\\x+1=\pm\sqrt{16-y}\\\\x=\pm\sqrt{16-y}-1$

Range of the function is defined as set of all values of y, for which x is defined.

⇒16 -y ≥ 0

⇒y ≤ 16

Option B

The domain is all real numbers. The range is {y|y ≤ 16}.

Andre45 [30]3 years ago

3 0

It is the 2nd one if im not right then idk

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A number, x, rounded to 1 decimal place is 3.7<br>Write down the error interval for x.

alekssr [168]

Answer:

$3.65 \leqslant x \leqslant 3.75$

Step-by-step explanation:

The error in rounding a number is half of the unit of measure.

The number was rounded to the nearest 0.1 unit so the error is 1/2×0.1, which equals 0.05.

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4 0

3 years ago

(08.07 HC)

andreev551 [17]

Answer:

$\textsf{A)} \quad x=-2, \:\:x=\dfrac{5}{2}$

$\textsf{B)} \quad \left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)$

C) See attachment.

Step-by-step explanation:

Given function:

$f(x)=2x^2-x-10$

To factor a <u>quadratic</u> in the form $ax^2+bx+c$ <em> , </em>find two numbers that multiply to $ac$ and sum to $b$ :

$\implies ac=2 \cdot -10=-20$

$\implies b=-1$

Therefore, the two numbers are -5 and 4.

Rewrite $b$ as the sum of these two numbers:

$\implies f(x)=2x^2-5x+4x-10$

Factor the first two terms and the last two terms separately:

$\implies f(x)=x(2x-5)+2(2x-5)$

Factor out the common term (2x - 5):

$\implies f(x)=(x+2)(2x-5)$

The x-intercepts are when the curve crosses the x-axis, so when y = 0:

$\implies (x+2)(2x-5)=0$

Therefore:

$\implies (x+2)=0 \implies x=-2$

$\implies (2x-5)=0 \implies x=\dfrac{5}{2}$

So the x-intercepts are:

$x=-2, \:\:x=\dfrac{5}{2}$

The x-value of the vertex is:

$\implies x=\dfrac{-b}{2a}$

Therefore, the x-value of the vertex of the given function is:

$\implies x=\dfrac{-(-1)}{2(2)}=\dfrac{1}{4}$

To find the y-value of the vertex, substitute the found value of x into the function:

$\implies f\left(\dfrac{1}{4}\right)=2\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)-10=-\dfrac{81}{8}$

Therefore, the vertex of the function is:

$\left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)$

Plot the x-intercepts found in Part A.

Plot the vertex found in Part B.

As the <u>leading coefficient</u> of the function is positive, the parabola will open upwards. This is confirmed as the vertex is a minimum point.

The axis of symmetry is the <u>x-value</u> of the <u>vertex</u>. Draw a line at x = ¹/₄ and use this to ensure the drawing of the parabola is <u>symmetrical</u>.

Draw a upwards opening parabola that has a minimum point at the vertex and that passes through the x-intercepts (see attachment).