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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 hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n
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Answer:

(-1,1),(4,-2)

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point (x,y)

Distance between points (x_1,y_1),(x_2,y_2) is given by \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]

Put (x,y)=(-1,1)

34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34

which is true. So, (-1,1) can be a vertex

Put (x,y)=(4,-2)

34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34

which is true. So, (4,-2) can be a vertex

Put (x,y)=(1,1)

34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22

which is not true. So, (1,1) cannot be a vertex

Put (x,y)=(2,-2)

34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22

which is not true. So, (2,-2) cannot be a vertex

Put (x,y)=(4,-1)

34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30

which is not true. So, (4,-1) cannot be a vertex

Put (x,y)=(-1,4)

34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70

which is not true. So, (-1,4) cannot be a vertex

So, possible points for the vertex are (-1,1),(4,-2)

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