A quadratic equation has x = -2 + 3i as a solution and passes through the point (-1, -15). Write the quadratic equation in standard form.

The standard form of the quadratic equation is f(x) = 2/3·(x + 2)² + 6

Step-by-step explanation:

The given solution of the quadratic equation is x = -2 + 3·i

A point on the path of the quadratic equation = (-1, -15)

The general form of the quadratic equation is f(x) = a·x² + b·x + c

When f(x) = 0, (At which the solution is found), we have x = -2 + 3·i

Substituting gives;

0 = a·(-2 + 3·i)² + b·(-2 + 3·i) + c

0 = 4·a - 12·a·i - 9·a - 2·b + 3·b·i + c

0 = 3·i·(b - 4·a) - 5·a - 2·b + c

∴ b = 4·a

c = 13·a

The general equation becomes

f(x) = a·x² + 4·a·x + 13·a

Also, when x = -1, f(x) = -15

f(-1) = -15 = a·(-1)² + 4·a·(-1) + 13·(-1)

-15 = a - 4·a - 13 = -3·a - 13

-3·a = -15 + 13 = -2

a = 2/3

The general form of the quadratic equation is therefore;

f(x) = 2/3·x² + 8/3·x + 13·2/3 = 2/3·x² + 8/3·x + 26/3

f(x) = 2/3·x² + 8/3·x + 26/3

The minimum value occurs at d(f(x)/dx = 0 = d(2/3·x² + 8/3·x + 13·2/3)/dx = 4/3·x + 8/3 = 0

x = -8/3 × 3/4 = -2

Therefore;

The minimum value of the quadratic function is f(-2) = 2/3·(-2)² + 8/3·(-2) + 13·2/3 = 6

The coordinates of the minimum point, which is the vertex point (h, k) = (-2, 6)

The standard form of the quadratic equation f(x) = a(x - h)² + k, is therefore;

f(x) = 2/3(x - (-2))² + 6 = 2/3·(x + 2)² + 6

f(x) = 2/3·(x + 2)² + 6

Also (h, k) can be gotten from h = -b/2a = -8/3/(2 × 2/3) = -2

k = c - b²/(4·a) = 26/3 - (8/3)²/(4 × 2/3) = 26/3 - 8/3 = 18/3 = 6

k = 6

The standard form of the quadratic equation f(x) = a(x - h)² + k becomes;

f(x) = 2/3·(x + 2)² + 6

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl