Question

i can write a quadratic equation in vertex form. i can graph a quadratic function. i can determine the axis of symmetry and vertex from a quadratic equation click here for a help video 7. complete the square to identify key parts of the following parabola and graph it on a coordinate plane, showing all work. $y = x^2 + 4x + 3$ a. vertex form of given equation: b. axis of symmetry: c. vertex: d. is the vertex a maximum or minimum: e. zero(s): f. y-intercept:

Explanation:

Step1: Complete the square for vertex form

Given \( y = x^2 + 4x + 3 \). To complete the square, take the coefficient of \( x \), which is 4, divide by 2 to get 2, and square it to get 4. Add and subtract 4 inside the equation:
\( y = (x^2 + 4x + 4) - 4 + 3 \)
Simplify the perfect square trinomial:
\( y = (x + 2)^2 - 1 \)

Step2: Find axis of symmetry

For a parabola in vertex form \( y = a(x - h)^2 + k \), the axis of symmetry is \( x = h \). From \( y = (x + 2)^2 - 1 \) (which can be written as \( y = (x - (-2))^2 + (-1) \)), \( h = -2 \), so the axis of symmetry is \( x = -2 \).

Step3: Find vertex

The vertex of \( y = a(x - h)^2 + k \) is \( (h, k) \). From \( y = (x + 2)^2 - 1 \), \( h = -2 \) and \( k = -1 \), so the vertex is \( (-2, -1) \).

Step4: Determine if vertex is max or min

The coefficient of \( x^2 \) in the original equation \( y = x^2 + 4x + 3 \) is 1 (positive), so the parabola opens upward. An upward-opening parabola has a minimum at its vertex.

Step5: Find zeros (x-intercepts)

Set \( y = 0 \) in the original equation: \( x^2 + 4x + 3 = 0 \). Factor the quadratic: \( (x + 1)(x + 3) = 0 \). Solve for \( x \): \( x + 1 = 0 \) or \( x + 3 = 0 \), so \( x = -1 \) or \( x = -3 \).

Step6: Find y-intercept

Set \( x = 0 \) in the original equation: \( y = 0^2 + 4(0) + 3 = 3 \).

Answer:

a. Vertex form: \( y = (x + 2)^2 - 1 \)
b. Axis of symmetry: \( x = -2 \)
c. Vertex: \( (-2, -1) \)
d. Minimum (since the parabola opens upward)
e. Zeros: \( x = -3 \) and \( x = -1 \)
f. y-Intercept: \( y = 3 \)