To graph the function \( f(x) = 4x^2 - 8x + 5 \) and find the vertex, let's first rewrite the function in vertex form.
Step 1: Identify the vertex
- The standard form of a quadratic function is \( f(x) = ax^2 + bx + c \). Here, \( a = 4 \), \( b = -8 \), and \( c = 5 \).
- The vertex \( x \)-coordinate can be found using the formula: \[ x = -\frac{b}{2a} = -\frac{-8}{2 \times 4} = \frac{8}{8} = 1 \]
- Substitute \( x = 1 \) into \( f(x) \) to find the \( y \)-coordinate of the vertex: \[ f(1) = 4(1)^2 - 8(1) + 5 = 4 - 8 + 5 = 1 \]
- The vertex is at the point \( (1, 1) \).
Step 2: Plot another point on the parabola
- Let's choose another value of \( x \) to find a point on the parabola. For example, let \( x = 0 \): \[ f(0) = 4(0)^2 - 8(0) + 5 = 5 \] So, the point \( (0, 5) \) is on the parabola.
Step 3: Plotting the points
- Plot the vertex \( (1, 1) \) on the graph.
- Plot the additional point \( (0, 5) \).
Step 4: Graph the parabola
- Since the coefficient of \( x^2 \) (which is 4) is positive, the parabola opens upwards.
- Connect the points smoothly to form a parabolic shape.
The result should show a vertex at \( (1, 1) \) and an additional point at \( (0, 5) \). The graph is symmetric around the vertical line \( x = 1 \).
Graph Summary
- Vertex: \( (1, 1) \)
- Another Point: \( (0, 5) \)
- Direction: Opens upwards
In a typical graph with \( x \) and \( y \) axes ranging from \( 0 \) to \( 10 \), the parabola would be nicely bound within those limits. The vertex would be at the bottom point of the curve.
If you were to sketch this out, you would illustrate the vertex at \( (1, 1) \), draw the curve opening upwards, and include the point \( (0, 5) \).
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