Locate the Vertex of the Quadratic Function f(x) = 3x^2 + 6x

Given the expression of the quadratic function

Finding the symmetry point of the function

$f(x)=3x^2+6x$

Video Solution

Solution Steps

00:00 Find the symmetry point in the function

00:03 The symmetry point is the point where if you fold the parabola in half

00:06 The halves will be equal to each other

00:10 We'll examine the function's coefficients

00:14 We'll use the formula to calculate the vertex point

00:18 We'll substitute appropriate values according to the given data and solve for X at the point

00:24 This is the X value at the symmetry point

00:28 Now we'll substitute this X value in the function to find Y value at the point

00:42 Negative squared always equals positive

00:55 This is the Y value at the symmetry point

00:59 And this is the solution to the question

Step-by-Step Solution

To find the symmetry point of the quadratic function $f(x) = 3x^2 + 6x$ , follow these steps:

Step 1: Identify the coefficients from the quadratic function $ax^2 + bx + c$ . Here, $a = 3$ and $b = 6$ .
Step 2: Use the vertex formula for the symmetry point, which is given by $x = -\frac{b}{2a}$ .
Step 3: Substitute the values of $a$ and $b$ into the formula:

$x = -\frac{6}{2 \times 3} = -\frac{6}{6} = -1$

Step 4: Substitute $x = -1$ back into the original function to find the corresponding $y$ -coordinate:

$f(-1) = 3(-1)^2 + 6(-1) = 3 \times 1 - 6 = 3 - 6 = -3$

Thus, the symmetry point of the given quadratic function $f(x) = 3x^2 + 6x$ is $\boxed{(-1, -3)}$ .