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aleksandr82 [10.1K]
3 years ago
15

What is the vertex of the parabola whose equation is y = (x + 1)2 + 3? (-1, -3) (-1, 3) (1, 3)

Mathematics
1 answer:
White raven [17]3 years ago
6 0
Algebra:
A standard parabola is y = x^2. Its vertex is at (0,0)
You can change the position (or vertex) of the parabola.
To move a parabola across the x-axis, you can add or subtract a number from x WITHIN brackets of the ^2
eg. (x + 1)^2 will move the parabola across the x-axis. It will move is one unit to the LEFT (as the sign is opposite to the direction it moves ie. The sign it + but you move the whole parabola in the -ve direction).
Adding or subtracting a number from x OUTSIDE of the ^2 moves the parabola up or down the y-axis
eg. x^2 + 3 will move the parabola UP 3 units (the sign is the same as the direction it moves when the added/subtracted number is outside of the ^2 ie. the sign is positive so the parabola moves up in the positive direction)

From this, we can conclude that because (x + 1)^2 + 3, the vertex will be where x = -1 and where y = 3

Vertex : (-1,3)



Calculus:
f(x) = (x + 1)^2 + 3 = x^2 + 2x + 1 + 3 = x^2 + 2x + 4
Expanding the formular to make it easier to differentiate

f'(x) = 2x + 2
Differentiating (finding the formular the the gradiet of the parabola)

0 = 2x + 2
When the gradient is equal to zero, it must be the vertex

-2 = 2x
-2/2 = x
x = -1
Solve to give the x value at the vertex

f(x) = (x + 1)^2 + 3
= (-1 + 1)^2 + 3
Substitute x = -1 into original equatiom to find y value at the vertex

= (0)^2 + 3
= 0 + 3
= 3
Solve for y

Vertex : (-1,3)


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The point (3, 6) is on the graph of y= 5f(2(x+3))-4 . Find the original point on the graph of y=f(x).
Sonja [21]

Answer:

<em>(12, 2)</em> is the original point on the graph of y=f(x).

Step-by-step explanation:

<u>Given:</u>

y= 5f(2(x+3))-4 has a point <em>(3, 6)</em> on its graph.

<u>To find:</u>

Original point on graph y=f(x) = ?.

<u>Solution:</u>

We are given that The point (3, 6) is on the graph of y= 5f(2(x+3))-4

If we put x = 3 and y = 6 in y= 5f(2(x+3))-4, it will satisfy the equation.

Let us the put the values and observe:

6= 5f(2(3+3))-4\\\Rightarrow 6= 5f(2(6))-4\\\Rightarrow 6= 5f(12)-4\\\Rightarrow 6+4=5f(12)\\\Rightarrow 5f(12)= 6+4\\\Rightarrow 5f(12)= 10\\\Rightarrow f(12)= \dfrac{10}{5}\\\Rightarrow f(12)= 2\\OR\\\Rightarrow 2=f(12)

Now, let us compare the above with the following:

y=f(x)

we get y = 2 and x = 12

So, the original point on graph of y=f(x) is <em>(12, 2).</em>

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