Question

Consider the quadratic function defined by:

Answer 1

Answer:

$\implies x = 1 \quad or -6 \\\\\implies f(x) = -3x^2-15x + 18$

Step-by-step explanation:

Given :-

• A quadratic function is given to us.
• The function is -3(x-1)(x-6)

And we need to find out the zeroes of the quadratic function and we need to express it in Standard form. T

Function :-

$\implies f(x)= -3(x-1)(x+6)$

• So for finding the zeroes equate it with 0 .So that ,

$\implies f(x)=0 \\\\\implies -3(x-1)(x+6)=0 \\\\\implies (x - 1 )(x+6) = 0 \\\\\implies x = 1 \quad or \quad -6$

Therefore the zeroes are 1 and -6 .

Expressing in Standard form :-

The standard form of a quadratic equation is ,

$\implies ax^2+bx + c = 0$

And that of a quadratic function is ,

$\implies f(x) = ax^2+bx + c$

Simplifying the equation :-

$\implies f(x)= -3(x-1)(x+6)\\\\\implies f(x) = (-3x +3)(x+6) \\\\\implies f(x) = -3x(x+6)+3(x+6)\\\\\implies f(x) = -3x^2 -18x +3x + 18\\\\\implies f(x) = -3x^2-15x + 18$

Hence the Standard form of the equation is -3x² - 15x + 18 .