Question

4 $y = x^2 + 2x + 4$
$y = x + 1$

Explanation:

Step1: Set equations equal

Since both equal \( y \), set \( x^2 + 2x + 4 = x + 1 \).

Step2: Rearrange into quadratic

Subtract \( x + 1 \): \( x^2 + x + 3 = 0 \).

Step3: Check discriminant

Discriminant \( D = b^2 - 4ac = 1^2 - 4(1)(3) = 1 - 12 = -11 \).

Step4: Analyze discriminant

Negative discriminant means no real solutions (no intersection).

Answer:

The system \( y = x^2 + 2x + 4 \) and \( y = x + 1 \) has no real - valued solutions (the graphs do not intersect at any real - valued points).