Question

1. the graph of a quadratic function has a vertex of (-2, -4) and passes through the origin. what is its equation written in standard form?

○ ( y = x^2 + 4x )
○ ( y = \frac{1}{8}x^2 + x )
○ ( y = -x^2 )
○ ( y = x^2 - 4x )

Explanation:

Step1: Recall vertex form of quadratic

The vertex form of a quadratic function is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. Given vertex \((-2, -4)\), substitute \( h = -2 \), \( k = -4 \):
\( y = a(x + 2)^2 - 4 \)

Step2: Use point (0,0) to find \( a \)

The graph passes through the origin \((0, 0)\). Substitute \( x = 0 \), \( y = 0 \) into the vertex form:
\( 0 = a(0 + 2)^2 - 4 \)
\( 0 = 4a - 4 \)
Solve for \( a \): \( 4a = 4 \Rightarrow a = 1 \)

Step3: Expand to standard form

Substitute \( a = 1 \) into vertex form:
\( y = (x + 2)^2 - 4 \)
Expand \((x + 2)^2\): \( x^2 + 4x + 4 \)
Thus, \( y = x^2 + 4x + 4 - 4 = x^2 + 4x \)

Answer:

\( y = x^2 + 4x \)