Questions: Draw a sketch of f(x)=-x^2+5. Plot the point for the vertex, and label the coordinate as a maximum or minimum, and draw and write the equation for the axis of symmetry. (1/2 point)

Draw a sketch of f(x)=-x^2+5. Plot the point for the vertex, and label the coordinate as a maximum or minimum, and draw and write the equation for the axis of symmetry. (1/2 point)

Transcript text: Draw a sketch of $f(x)=-x^{2}+5$. Plot the point for the vertex, and label the coordinate as a maximum or minimum, and draw \& write the equation for the axis of symmetry. ( $1 / 2$ point)

Solution

Solution Steps

Step 1: Identify the Function

The given function is $ f(x) = -x^2 + 5 $.

Step 2: Determine the Vertex

The function $ f(x) = -x^2 + 5 $ is a quadratic function in the form $ f(x) = ax^2 + bx + c $ where $ a = -1 $, $ b = 0 $, and $ c = 5 $. The vertex of a parabola in this form is given by the formula: \[ x = -\frac{b}{2a} \] Substituting the values of $ a $ and $ b $: \[ x = -\frac{0}{2(-1)} = 0 \] The y-coordinate of the vertex is: \[ f(0) = -(0)^2 + 5 = 5 \] Thus, the vertex is at $ (0, 5) $.

Step 3: Determine the Axis of Symmetry

The axis of symmetry for a parabola in the form $ f(x) = ax^2 + bx + c $ is the vertical line $ x = -\frac{b}{2a} $. From Step 2, we found this to be $ x = 0 $.

Step 4: Determine if the Vertex is a Maximum or Minimum

Since the coefficient of $ x^2 $ is negative ($ a = -1 $), the parabola opens downwards, and the vertex represents a maximum point.

Final Answer

The vertex of the function is at $ (0, 5) $ and it is a maximum.
The axis of symmetry is the line $ x = 0 $.

{"axisType": 3, "coordSystem": {"xmin": -3, "xmax": 3, "ymin": -1, "ymax": 6}, "commands": ["y = -x**2 + 5"], "latex_expressions": ["$y = -x^2 + 5$"]}

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