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krek1111 [17]
2 years ago
10

What is the vertex of the graph of the following equation? y = x^2 + 2x-3

Mathematics
1 answer:
ExtremeBDS [4]2 years ago
6 0

Answer:

vertex = (- 1, - 4 )

Step-by-step explanation:

Given a parabola in standard form

y = ax² + bx + c ( a ≠ 0 )

Then the x- coordinate of the vertex is

x = - \frac{b}{2a}

y = x² + 2x - 3 ← is in standard form

with a = 1, b = 2 , then

x = - \frac{2}{2} = - 1

Substitute x = - 1 into the equation for corresponding value of y

y = (- 1)² + 2(- 1) - 3 = 1 - 2 - 3 = - 4

vertex = (- 1, - 4 )

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Which expression is a difference of squares with a factor of 2x5?
Sedbober [7]

Answer:

{( \sqrt{3.5} })^{2}  -  {( \sqrt{1.5} )}^{2}

Step-by-step explanation:

The difference of two squares identity is

{x}^{2}   -  {y}^{2}  = (x  - y)(x + y)

When we let

\sqrt{3.5}

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{( \sqrt{3.5} )}^{2}  -  {( \sqrt{1.5}) }^{2}  = (3.5 - 1.5)(3.5 + 1.5)

This will simplify to give us the required factors

{( \sqrt{3.5} )}^{2}  -  {( \sqrt{1.5}) }^{2}  =2 \times 5

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3 years ago
The point P(8,-3) lies on the curve with equation y=f(x). Find the image of P on the curve with equation y=f(x-4).
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Answer:

(12, -3)

Step-by-step explanation:

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