Thread: Q from From deeqow

Deeqow has PM me this question I am posting it here for anyone who can help.

Q: The equation of a parabola is given by Y=x2+2x-4 and the straight line by Y=x+2.
find the point where the parabola crosses x-axis.
find the point where the parabola crosses y-axis.
find the parabola axis of symmetry.
find the coordinates the parabola's minimum point.
find the point where the parabola crosses the straight line.

1. Find the point where Y=x2+2x-4 crosses x axis:

When y = 0 the parabola is on the x axis so solve 0=x^2+2x-4
This equation is of the form ax^2 + bx + c
a=1
b=2
c=-4

To find the roots we use the quadratic equation:

plug and chug and you get -1+sqrt(5) and -1-sqrt(5)

2. To find the point where Y=x^2+2x-4 crosses the y axis we need to find the solution to the equation when x = 0:

Y=0^2+2*0-4
Y=-4

Here is the graph

To find the parabola's axis of symmetry and the minumum point you need to either use a graphing calculator to find the min on the curve or take the derivative and set it equal to zero because the slope at the min is always zero.

Derivative of x^2 + 2x - 4 = 2x+2

2x +2 = 0

x=-1 this is the minimum point

The parabola's axis of symmetry is a vertical line through this point which can be expressed as x=-1

The axis of symmetry can be found more easily by x=-b/2a, the vertex or minimum point is obviously (-b/2a,f(-b/2a)), the points of intersection can be found by using the substitution method and then factoring.

y = x^2 +2x -4 ....(1)
Because it is the x that is squared, and the x^2 is positive, then this is a vertical parabola that opens upward.

A standard form of a vertical parabola that opens upward is
(y-k) = a(x-h)^2 ....(2)
where
(h,k) is the vertex or lowest point.
a is any real number.

Let us transform (1) into (2).
y = x^2 +2x -4 ....(1)
We complete the square of the x terms,
y +4 = x^2 +2x
y +4 = x^2 +2x +(2/2)^2 -(2/2)^2
y +4 = x^2 +2x +1 -1
y +4 +1 = x^2 +2x +1
y +5 = (x+1)^2 ....(3)
There, the vertical parabola in standard form.

That means k = -5; h = -1, a=1
Or the lowest point or vertex is (h,k) = (-1,-5).

-------------------
>>>find the axis of symmetry.

The axis of symmetry of a vertical parabola is a vertical line that passes through the vertex.
Therefore, the axis of symmetry is the line x = -1. ....answer.

------------
>>>Find the coordinates of the parabola's minimum point.

The mimimum point of a vertical parabola is its lowest point or vertex.
Therefore, the minimum point is (-1,-5). ....answer.

-------------
>>>find the point where the parabola crosses x-axis.

When the parabola crosses the x-axis, the y-value is zero, or y=0.
So,
y +5 = (x+1)^2 ....(3)
0 +5 = (x+1)^2
(x+1)^2 = 5
Take the square roots of both sides,
(x+1) = +,-sqrt(5)
Or,
x = -1 +sqrt(5) ,or about 1.236
x = -1 -sqrt(5) , or about -3.236

That means there are 2 points where the parabola crossses the x-axis:
(-3.236,0) and (1.236,0) ....answer.

------------
>>>find the point where the parabola crosses y-axis.

When the parabola crosses the y-axis, the x-value is zero, or x=0.
So,
y +5 = (x+1)^2 ....(3)
y +5 = (0+1)^2
y +5 = (1)^2
y +5 = 1
y = 1 -5
y = -4

That means the parabola crossses the y-axis at only one point:
(0,-4) ....answer.

---------------
>>>find the point where the parabola crosses the straight line.

When the parabola crosses the the straight line y = x+2, at the points of intersection the parabola and the straight line have the same x- and y-values.
So,
y = x+2
Substitute that in (1),
y = x^2 +2x -4 ....(1)
x+2 = x^2 +2x -4
0 = x^2 +2x -x -4 -2
0 = x^2 +x -6
Or,
x^2 +x -6 = 0
Factor that,
(x+3)(x-2) = 0

So,
x+3 = 0
x = -3

x-2 = 0
x = 2

When x = -3,
y = x+2 = -3+2 = -1
That is point (-3,-1)

when x = 2,
y = x+2 = 2+2 = 4
That is point (2,4)

Therefore, the the parabola crosses the straight at points (-3,-1) and (2,4). ....answer.