Question

graph the function.
$y = x^2 + 6x + 5$

Explanation:

Step1: Find the vertex form

To graph \( y = x^2 + 6x + 5 \), we complete the square. The formula for completing the square for \( ax^2+bx+c \) is \( a(x - h)^2 + k \), where \( h=-\frac{b}{2a} \) and \( k = c-\frac{b^2}{4a} \). For \( a = 1 \), \( b = 6 \), \( c = 5 \), \( h=-\frac{6}{2\times1}=-3 \), \( k = 5-\frac{6^2}{4\times1}=5 - 9=-4 \). So the vertex form is \( y=(x + 3)^2-4 \). The vertex is \((-3,-4)\).

Step2: Find the x - intercepts

Set \( y = 0 \), so \( x^2+6x + 5=0 \). Factor the quadratic: \( (x + 1)(x + 5)=0 \). Solving gives \( x=-1 \) or \( x=-5 \). So the x - intercepts are \((-5,0)\) and \((-1,0)\).

Step3: Find the y - intercept

Set \( x = 0 \), then \( y=0^2+6\times0 + 5 = 5 \). So the y - intercept is \((0,5)\).

Step4: Plot the points and draw the parabola

Plot the vertex \((-3,-4)\), x - intercepts \((-5,0)\), \((-1,0)\), and y - intercept \((0,5)\). Since \( a = 1>0 \), the parabola opens upwards. Draw a smooth curve through these points.

Answer:

To graph \( y=x^{2}+6x + 5 \):

1. Vertex: \((-3,-4)\) (from vertex form \( y=(x + 3)^2-4 \)).
2. x - intercepts: \((-5,0)\) and \((-1,0)\) (from factoring \( (x + 1)(x + 5)=0 \)).
3. y - intercept: \((0,5)\) (from \( x = 0 \)).
4. Plot these points and draw a upward - opening parabola through them.