Question

Identify the graph of g(x)=4x2+24x+38




HINT

Quadratic functions - Standard form Lesson Example ,Video and Practice
Quadratic functions - Vertex form Lesson Example ,Video and Practice
Quadratic functions from x intercepts and points
(1 point)
Responses

[A]
[A]

[B]
[B]

[C]
[C]

[D]

Answer 1

To identify the graph of the quadratic function \( g(x) = 4x^2 + 24x + 38 \), we can analyze its characteristics.

1. **Standard Form**: The function is in standard form \( g(x) = ax^2 + bx + c \), where:
- \( a = 4 \) (positive, so the parabola opens upwards)
- \( b = 24 \)
- \( c = 38 \)

2. **Vertex**: The vertex \( x \)-coordinate can be found using the formula \( x = -\frac{b}{2a} \).
\[
x = -\frac{24}{2 \cdot 4} = -\frac{24}{8} = -3
\]
To find the \( y \)-coordinate of the vertex, substitute \( x = -3 \) back into the function:
\[
g(-3) = 4(-3)^2 + 24(-3) + 38 = 4(9) - 72 + 38 = 36 - 72 + 38 = 2
\]
Thus, the vertex is at the point \((-3, 2)\).

3. **Y-intercept**: The \( y \)-intercept occurs when \( x = 0 \):
\[
g(0) = 4(0)^2 + 24(0) + 38 = 38
\]
So the \( y \)-intercept is \( (0, 38) \).

4. **X-intercepts**: To find the x-intercepts (if any), we can set \( g(x) = 0 \) and solve for \( x \):
\[
4x^2 + 24x + 38 = 0
\]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( b^2 - 4ac = 24^2 - 4 \cdot 4 \cdot 38 = 576 - 608 = -32 \)
Since the discriminant is negative, there are no real x-intercepts.

5. **Shape of the graph**: Since \( a > 0 \), the parabola opens upwards and has no x-intercepts.

In summary:
- The vertex is at \((-3, 2)\).
- The y-intercept is at \( (0, 38) \).
- There are no x-intercepts.

Given these characteristics, you can match them with the choices provided in your responses to identify the correct graph. If you have the graphs available, look for a parabola that opens upwards, has a vertex at (-3, 2), and intersects the y-axis at (0, 38).