Question

Graph the quadratic y = 4x² - 5
*
1 point
(-1, 0)
(1, 0)
(0, 4)
(0, -5)

Answer 1

To graph the quadratic equation \( y = 4x^2 - 5 \), we can first identify its key features:

1. **Vertex**: The vertex of the parabola can be found by looking at the standard form of a quadratic equation, which is \( y = ax^2 + bx + c \). For our equation, \( a = 4 \), \( b = 0 \), and \( c = -5 \). The vertex is found at \( x = -\frac{b}{2a} = -\frac{0}{2 \cdot 4} = 0 \). Substituting \( x = 0 \) back into the equation gives us \( y = 4(0)^2 - 5 = -5 \). So the vertex is at the point \( (0, -5) \).

2. **Y-intercept**: To find the y-intercept, we can substitute \( x = 0 \) into the equation, which gives us the same result: \( y = -5 \). Thus, the y-intercept is also \( (0, -5) \).

3. **X-intercepts**: To find the x-intercepts, we set \( y = 0 \):
\[
0 = 4x^2 - 5
\]
Solving for \( x \):
\[
4x^2 = 5
\]
\[
x^2 = \frac{5}{4}
\]
\[
x = \pm \frac{\sqrt{5}}{2}
\]
This means our x-intercepts are approximately \( (-1.12, 0) \) and \( (1.12, 0) \).

From the points you've provided:
- \( (-1, 0) \) (approximate to one of the x-intercepts)
- \( (1, 0) \) (approximate to one of the x-intercepts)
- \( (0, 4) \) (not on the graph)
- \( (0, -5) \) (the y-intercept)

The correct points to plot on the graph would be:
- \( (0, -5) \) as the vertex and y-intercept
- Approximate x-intercepts at \( (-1.12, 0) \) and \( (1.12, 0) \)

To summarize:
- To graph the quadratic \( y = 4x^2 - 5 \), plot the points \( (0, -5) \) and the x-intercepts, and then draw a parabola opening upward through these points. The selected points you accurately identified on the provided list are \( (0, -5) \) and approximated \( (1, 0) \) and \( (-1, 0) \).

The key point for the quadratic graph is \( (0, -5) \).