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3 years ago

Passing through (1,9),slope=2

Mathematics

2 answers:

Natalka [10]3 years ago

5 0

Passes through (1, 9) with a slope of 2.

Assuming that we want this in slope intercept form, which is y=mx+b (m=slope, b=y-intercept) then we need to find the y-intercept! :)

To find the y-intercept (or b!!) you simply need to plug everything into slope intercept form using our given points.

y = mx + b

Plug everything in.

9 = 2(1) + b

Simplify.

9 = 2 + b

Subtract 2 from both sides.

9 - 2 = b

b = 7

Therefore, our y-intercept is 7! :)

Now that we have both the y-intercept and the slope, we can finally plug this into a new equation, using slope intercept form.

y = mx + b

y = 2x + 7

~Hope I helped!~

Bumek [7]3 years ago

3 0

y = ax + b

Here the slope is 2.

Then, y = 2x + b

The line is passing through (1;9).

We just have to substitute that in the equation.

9 = 2*1+ b
b = 9 - 2
b = 7

That's the intercept.

Eventually, the equation is : y = 2x + 7

Hope this helps !

Photon

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2 years ago

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algol13

Answer:

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General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Factoring

Calculus

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

Step 2: Differentiate Pt. 1

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Implicit Differentiation [Basic Power Rule]: $-y'-6x^2=2yy'$
[Algebra] Isolate y' terms: $-6x^2=2yy'+y'$
[Algebra] Factor y': $-6x^2=y'(2y+1)$
[Algebra] Isolate y': $\frac{-6x^2}{(2y+1)}=y'$
[Algebra] Rewrite: $y' = \frac{-6x^2}{(2y+1)}$

Step 3: Differentiate Pt. 2

Find 2nd Derivative

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}$
[Derivative] Back-Substitute y': $y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}$

Step 4: Find Slope at Given Point

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[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}$
[Pre-Algebra] Divide: $y''(-1,-2) = \frac{-36+24 }{9}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-12}{9}$
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dimaraw [331]

The answer would be 28.
--
Here's my work:

Remember PEMDAS:
- Parentheses
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Also, remember to always go left to right.

(6+2)^2-4*3
Square everything in the parenthesis.

36+4-4*3
Multiply where you see a multiplication symbol from left to right.

36+4-12
Add.

40-12
Subtract.

28
Done!
--
I hope I was able to guide and help you on this.

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3 years ago

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GalinKa [24]

Answer:

X- intercept(S) : (4,0)

Y- Intercept(S) : (0, -6)

Step-by-step explanation:

To find the x-intercept, substitute in 0 for y and solve for x. To find the y-intercept, substitute in 0 for x and solve for Y.

Hopes this helps!

8 0

2 years ago