Standard form of equation
The equations are named based on the degree of the equations.
The equations with degree 1 are known as linear equations.
The equations of degree 2 are known as quadratic equations.
The equations of degree 3 are known as cubic equations and so on.
Standard form of linear equation
The standard form of a linear equation is given by Ax+ By = C where A, B and C are constants and x and y are variables
Find the equation of the line passing through (6, -2) and having a slope 3.
We know that if a line passes through a point (x1, y1) and have slope m, then the equation is given by y - y1 = m (x - x1). This form is known as point slope form.
Given that the line passing through (6, -2). That is (x1, y1) = (6, -2).
Slope is 3. That is m = 3
Therefore the required line is
y - (-2) = 3 (x - 6)
y + 2 = 3x - 18
adding 18 both sides, we get
y + 20 = 3x
subtracting y on both sides, we get
3x - y = 20, which is the required equation.
Standard form of quadratic equation
The standard form of the quadratic equation is ax2 + bx + c = 0, a ≠ 0.
Write the quadratic equation in standard form.
2x2 = 12x + 3
We know that the standard form of the quadratic equation is given by ax2 + bx + c = 0, a ≠ 0.
Given equation is 2x2 = 12x + 3
To convert this to standard form, we take all terms to left.
Subtracting 12x and 3 on both sides, we get
2x2 - 12x - 3 = 0, which is the required standard form
Standard form of conic sections
A conic section or conic is defined as the locus of a point which moves in a plane so that its distance from a fixed point in the plane bears a constant ratio to its distance from a fixed line in the plane. The fixed point is called focus, the fixed line is called the directrix, and the constant ratio is called eccentricity of the conic. The eccentricity of a conic is denoted by e
Standard form of a parabola
The standard form of a parabola is given by y = ax2 + bx + c, a ≠ 0. It is the general quadratic equation.
If a > 0, the parabola opens upwards and if a < 0, the parabola opens downwards.
Write the equation of the parabola whose focus is S(-4,5) and vertex (-4,1).
Distance between the focus and the vertex = 4
Distance between the vertex and the directrix = 4
Distance between the vertex and x axis = 1
Therefore distance between x axis and directrix =3
Since the directrix is parallel to x axis and at a distance of 3 units below the x axis, the equation to the directrix is y = -3
Let P(x,y) be any point on the parabola. Let PM be the perpendicular to the directrix.
Then SP2 = (x + 4)2 + (y - 5)2
PM2 = (y + 3)2
Since SP2 = PM2, we get
(x + 4)2 + (y - 5)2 = (y + 3)2
Simplifying we get
x2 + 8x + 16 + y2 - 10y + 25 = y2 + 6y + 9
Cancelling common terms, we get
x2 + 8x + 16 - 10y + 25 = 6y + 9
subtracting 6y and 9 on both sides, we get
x2 + 8x -16y +32 =0
Write all terms in x and constants on one side
16y = x2 + 8x + 32
Dividing by 16 on both sides, we get
y = x2/16 + 1/2x + 2, is the required equation of the parabola.
Standard form of a circle
If the center of the circle is at the point (h, k) and has radius of the circle is r, then the equation of the circle is given by (x - h)2+ (y - k)2= r2.
Write the equation of the circle in the standard form, whose center is (6, 9) and radius is 2.
Given center of the circle,(h, k) = (6, 9) and radius of the circle, r = 2.
We know that the standard form of a circle is (x - h)2 + (y - k)2 = r2
Substituting for h, k and r, we get
(x - 6)2 + (y - 9)2 = 22
Simplifying we get,
(x - 6)2 + (y - 9)2 = 4
So the required equation is (x - 6)2 + (y - 9)2 = 4.
Standard form of numbers
Write the number 47956.759 in the standard form.
Given number is 47956.759. we have to write this number in standard form.
In the standard form we write the number in the form x * 10y,where x is a decimal number with only one number before decimal.
That is 47956.759 = 4.7956759 x 104
So the standard form of 47956.759 = 4.7956759 x 104
Converting standard form
Converting parabola from standard from to vertex form
To convert standard form to vertex form we follow the following procedure: -
- First we have to take the common factors from the first two terms in the standard form f(x) = ax2 + bx + c
- Now we complete the square of the first two terms by adding and subtracting square of the half of second term.
- Simplify and rearrange it in the form y = a(x - h)2 + k
Convert the parabola f(x) = x2 - 6x + 6 in the standard form to vertex form and hence find the vertex.
Given standard form of parabola is f(x) = x2 - 6x + 6.
Comparing with the standard form of parabola, f(x) = ax2 + bx + c, we get
a = 1.
Since a = 1, we complete the square of the first two terms in the quadratic equation. To complete the square, we add and subtract the square of the half of the second term.
So we get
f(x) = x2 - 6x + 9 - 9 + 6
Now we can combine the first three terms and last two term.
f(x) = (x - 3)2 - 3, which is the required vertex form and the vertex is given by (3, -3).
Converting slope intercept form to standard form
When we take all the variables to one side and keep the constants on the other side, we get the standard form.
Convert the line y = 5x + 8 into standard form.
Given line is y = 5x + 8. To convert into standard form, we have to take all variables to one side, keeping the constants on the other side.
For this lets subtract 5x from both sides,
y = 5x + 8
y - 5x = 8
So the required equation is -5x + y = 8
Functions in standard form
Standard form of a quadratic function
The standard form of a quadratic function is given by
f(x) = a(x - h)2 + k, where a ≠ 0.
This form is also known as vertex form of a quadratic function
Standard form of a polynomial function
When we arrange the polynomial in the decreasing order of its degree, it is said to be in the standard form. In general, an nth degree polynomial P(x) = anxn + an-1xn-1 + ...+ a1x + a0 where an, an-1,... are constants and an ≠ 0 is in the standard form. Here we can see that the power of x in each term decreases.