
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+2x4 and the straight line by Y=x+2.
find the point where the parabola crosses xaxis.
find the point where the parabola crosses yaxis.
find the parabola axis of symmetry.
find the coordinates the parabola's minimum point.
find the point where the parabola crosses the straight line.

Part one
1. Find the point where Y=x2+2x4 crosses x axis:
When y = 0 the parabola is on the x axis so solve 0=x^2+2x4
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: http://mathworld.wolfram.com/qimg560.gif
plug and chug and you get 1+sqrt(5) and 1sqrt(5)

part 2
2. To find the point where Y=x^2+2x4 crosses the y axis we need to find the solution to the equation when x = 0:
Y=0^2+2*04
Y=4

1 Attachment(s)

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
(yk) = a(xh)^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 xaxis.
When the parabola crosses the xaxis, the yvalue 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 xaxis:
(3.236,0) and (1.236,0) ....answer.

>>>find the point where the parabola crosses yaxis.
When the parabola crosses the yaxis, the xvalue 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 yaxis 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 yvalues.
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)(x2) = 0
So,
x+3 = 0
x = 3
x2 = 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.