Question

graph the function f(x) = 4x². plot the vertex. then plot another point on the parabola. if you make a mistake, you can erase your parabola by selecting the second point and placing it on top of the first.

Explanation:

Step1: Find the vertex of the parabola

For a quadratic function in the form \( f(x) = ax^2 + bx + c \), the vertex is at \( x = -\frac{b}{2a} \). In \( f(x) = 4x^2 \), \( a = 4 \), \( b = 0 \), so \( x = -\frac{0}{2\times4} = 0 \). Substitute \( x = 0 \) into the function: \( f(0) = 4\times0^2 = 0 \). So the vertex is \( (0, 0) \).

Step2: Find another point on the parabola

Let's choose \( x = 1 \). Substitute \( x = 1 \) into \( f(x) = 4x^2 \): \( f(1) = 4\times1^2 = 4 \). So the point \( (1, 4) \) is on the parabola. (We could also choose \( x = -1 \), and \( f(-1) = 4\times(-1)^2 = 4 \), so \( (-1, 4) \) is also on the parabola, due to symmetry.)

Step3: Plot the points

Plot the vertex \( (0, 0) \) and the point \( (1, 4) \) (or \( (-1, 4) \)) on the coordinate plane. Then draw the parabola opening upwards (since \( a = 4 > 0 \)) passing through these points.

Answer:

The vertex is \( (0, 0) \) and another point (e.g., \( (1, 4) \) or \( (-1, 4) \)) can be plotted to graph the parabola \( f(x) = 4x^2 \).