Question

1. y = x² - 2x + 5

vertex form:
vertex

1. y = x² + 4x - 3

vertex form:
vertex

Explanation:

Step1: Complete the square for \(y = x^{2}-2x + 5\)

We know that \((a - b)^2=a^{2}-2ab + b^{2}\). For \(y=x^{2}-2x + 5\), in the terms \(x^{2}-2x\), if \(a = x\) and \(-2ab=-2x\), then \(b = 1\). So \(y=(x^{2}-2x+1)+4=(x - 1)^{2}+4\).

Step2: Identify the vertex

For a quadratic function in vertex - form \(y=a(x - h)^{2}+k\), the vertex is \((h,k)\). Here \(a = 1\), \(h = 1\) and \(k = 4\), so the vertex is \((1,4)\).

Step3: Complete the square for \(y=x^{2}+4x-3\)

For the terms \(x^{2}+4x\), if \(a = x\) and \(2ab = 4x\), then \(b = 2\). So \(y=(x^{2}+4x + 4)-4-3=(x + 2)^{2}-7\).

Step4: Identify the vertex

For the quadratic function \(y=(x + 2)^{2}-7\) in vertex - form \(y=a(x - h)^{2}+k\) (where \(a = 1\), \(h=-2\) and \(k=-7\)), the vertex is \((-2,-7)\).

Answer:

For \(y=x^{2}-2x + 5\), Vertex Form: \(y=(x - 1)^{2}+4\), Vertex: \((1,4)\)
For \(y=x^{2}+4x-3\), Vertex Form: \(y=(x + 2)^{2}-7\), Vertex: \((-2,-7)\)