To analyze the quadratic function $y = -x^2 + 8x - 10$, let’s identify the following:

1. Axis of Symmetry
2. Vertex
3. y-intercept
4. Range

Step 1: Axis of Symmetry

For a quadratic function in the form $y = ax^2 + bx + c$, the axis of symmetry is given by: $x = -\frac{b}{2a}$ In this function, $a = -1$ and $b = 8$. Substituting these values: $x = -\frac{8}{2(-1)} = \frac{8}{2} = 4$ So, the axis of symmetry is $x = 4$.

Step 2: Vertex

The vertex lies on the axis of symmetry. To find the y-coordinate of the vertex, substitute $x = 4$ into the equation: $y = -(4)^2 + 8 \cdot 4 - 10$ $y = -16 + 32 - 10$ $y = 6$ Thus, the vertex is at $(4, 6)$.

Step 3: y-intercept

The y-intercept is the value of $y$ when $x = 0$: $y = -(0)^2 + 8 \cdot 0 - 10 = -10$ So, the y-intercept is $(0, -10)$.

Step 4: Range

Since the coefficient of $x^2$ (i.e., $a = -1$) is negative, the parabola opens downwards. The maximum value of $y$ is at the vertex, $y = 6$. Therefore, the range of the function is: $(-\infty, 6]$

Summary

• Axis of Symmetry: $x = 4$
• Vertex: $(4, 6)$
• y-intercept: $(0, -10)$
• Range: $(-\infty, 6]$

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Related Questions

1. How do you determine if a parabola opens upwards or downwards?
2. How can you find the x-intercepts of a quadratic function?
3. What is the domain of any quadratic function?
4. How do you convert a quadratic function to vertex form?
5. How can completing the square help find the vertex?

Tip:

Remember, for a quadratic function $y = ax^2 + bx + c$, if $a > 0$, the parabola opens upwards; if $a < 0$, it opens downwards.