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4 years ago

What are the coordinates of the vertex for f(x) = 2x2 + 4x + 9?

Mathematics

2 answers:

KatRina [158]4 years ago

8 0

Answer:

The answer to your question is V ( -1 , 7)

Step-by-step explanation:

Data

f(x) = 2x² + 4x + 9

1.- Use "y" instead of f(x)

y = 2x² + 4x + 9

2.- Subtract 9 in both sides and simplify

y - 9 = 2x² + 4x + 9 - 9

y - 9 = 2x² + 4x

3.- Factor 2 in the right side

y - 9 = 2(x² + 2x )

4.- Complete the perfect square trinomial

y - 9 + 2 = 2(x² + 2x + 1)

5.- Simplify

y - 7 = 2(x + 1)²

6.- Get the Vertex

V ( -1 , 7)

Scorpion4ik [409]4 years ago

5 0

To answer this quickly the answer is (-1,7)

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Archy [21]

Answer:

D) $0 Per Month.

Step-by-step explanation:

In January, Mr. Sanders spent $50 on gas. In May, Mr. Sanders spent $50 on gas. The amount that he spent on gas did not change from January to May, so the rate of change between January and May is D) $0 Per Month.

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4 years ago

How to graph this exponential function

Alexeev081 [22]

X          f(x)
----       -------------------------------
-2         4(1/2)⁻² = 4.[1/(1/2)²] = 16
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3 years ago

Determine the value of k for which -3x^2+12x+k=0 has no real roots.

Crazy boy [7]

Answer:

K<-12

Step-by-step explanation:

+ For that, we need the delta negatif.

+ The delta= b^2-4ac=12^2-4(-3)(k)=144+12k<0

Then, 12k<-144 (divise both sides by 12)

we have k<-12

Or an other way:

-3x^2+12x+k=0 (divise both sides by -3)

x^2-4x-k/3=0

x^2-4x+4 -4-k/3=0

(x-2)^2=4+k/3

Now , we need 4+k/3<0, k/3<-4 or k<-12

Hope you understand.

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3 years ago

One side of a triangle is increasing at a rate of 9 cm/s and a second side is decreasing at a rate of 2 cm/s. if the area of the

LenKa [72]

Use the attached figure to guide yourself through the explanation. In the figure:
$a=21cm, b=36cm, \widehat{C}=\frac{\pi}{6}$ .
We know that:
$\frac{da}{dt}=9,\frac{db}{dt}=-2$
Also remmeber that the area of a triangle is:
$A=\frac{Base\times Height}{2}$
The hegiht is shown in the figure in red, the base is just the b side.
Using the angle C the area of the triangle is:
$A=\frac{1}{2}ab\cos(\widehat{C})\implies2A=ab\cos(\widehat{C})$
Now taking the derivative of the area 2A with respect to t we have:
$2\frac{dA}{dt}=b\cos(\widehat{C})\frac{da}{dt}+a\cos(\widehat{C})\frac{db}{dt}-ab\sin(\widehat{C})\frac{d\widehat{C}}{dt}$

Replacing with the right values:
$\frac{dA}{dt}=0\\
\\
\cos({\widehat{C}})=\frac{\sqrt{3}}{2}\\ \\
\sin({\widehat{C}})=\frac{1}{2}\\ \\

$
yields:
$0=36\frac{\sqrt{3}}{2}.9-21\frac{\sqrt{3}}{2}.2-\frac{1}{2}.21.36.\frac{d\widehat{C}}{dt}\implies \frac{d\widehat{C}}{dt}=\frac{141}{378}\sqrt{3}$