Best solutions to - "How to draw the parabola y=3x(x+2)?"

Bentley Doyle

Answered question

2023-01-06

How to draw the parabola

y = 3 x (x + 2)

Aryanna Chaney

Beginner2023-01-07Added 6 answers

y = 3 x (x + 2)

Distribute the 3x into the parenthesis

y = 3 x^{2} + 6 x

For the quadratic

y = a x^{2} + b x + c

the vertex (h, k) which is the same as (x,y) is found by solving for

h = - \frac{b}{2 a}

y = 3 x^{2} + 6 x

a = 3

b = 6

c = 0

h = - \frac{6}{2 (3)} = - \frac{6}{6} = - 1

h = - 1

(h, k) \Rightarrow (- 1, k)

Now, enter the discovered h value (-1) into the initial equation to determine the value of k (your y value)

y = 3 x^{2} + 6 x

y = 3 {(- 1)}^{2} + 6 (- 1)

y = 3 (1) - 6

y = 3 - 6

y = - 3

(h, k) \Rightarrow (- 1, - 3)

(- 1, - 3)

is the vertex of the equation

Keely Rivers

Beginner2023-01-08Added 1 answers

I wanted to give a quick derivation of why the vertex point is at

0 x = - \frac{b}{2 a}

for anyone who understands differential calculus:
The vertext point, or stationary point, is the location where the function's gradient is zero.
Let

y = a x^{2} + b x + c

Using power rule:

\frac{d y}{d x} = 2 a x + b

When gradient is 0:

2 a x + b = 0

\Rightarrow 2 a x = - b

\Rightarrow x = - \frac{b}{2 a}

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