Questions: Match the function with its graph. y=5(x+2)^2-1

Solution Steps

The given function is \(y = 5(x+2)^2 - 1\). This is a quadratic function in vertex form, where the vertex is given by \((-h, k)\). In this case, \(h = -2\) and \(k = -1\), so the vertex is \((-2, -1)\). The coefficient of the \((x+2)^2\) term is 5, which is positive, meaning the parabola opens upwards.

We are looking for a parabola that opens upwards with a vertex at \((-2, -1)\). The first graph has a vertex at \((1,0)\). The second graph has a vertex at \((1,0)\). The third graph has a vertex at \((-1,0)\). The fourth graph has the vertex at \((-2, -1)\) and it opens upwards.

Final Answer