Question

find the y - intercept, the axis of symmetry, and the vertex of the graph of the function.
$f(x) = 0.4x^2 + 2.4x$
the y - intercept is $(0,0)$. (type an ordered pair.)
the axis of symmetry is $x = 0$. (simplify your answer. type an equation.)

Explanation:

Step1: Find the y - intercept

To find the y - intercept, we set \(x = 0\) in the function \(f(x)=0.4x^{2}+2.4x\).
Substitute \(x = 0\) into the function: \(f(0)=0.4(0)^{2}+2.4(0)=0\). So the y - intercept is \((0,0)\).

Step2: Find the axis of symmetry

For a quadratic function in the form \(f(x)=ax^{2}+bx + c\), the formula for the axis of symmetry is \(x=-\frac{b}{2a}\).
In the function \(f(x)=0.4x^{2}+2.4x\), we have \(a = 0.4\) and \(b = 2.4\).
Substitute \(a\) and \(b\) into the formula: \(x=-\frac{2.4}{2\times0.4}=-\frac{2.4}{0.8}=- 3\).

Step3: Find the vertex

The x - coordinate of the vertex is the same as the axis of symmetry, so \(x=-3\).
To find the y - coordinate of the vertex, substitute \(x = - 3\) into the function \(f(x)=0.4x^{2}+2.4x\).
\(f(-3)=0.4(-3)^{2}+2.4(-3)=0.4\times9-7.2 = 3.6 - 7.2=-3.6\). So the vertex is \((-3,-3.6)\).

Answer:

• Y - intercept: \((0,0)\)
• Axis of symmetry: \(x=-3\)
• Vertex: \((-3,-3.6)\)