To determine the coordinates of $P$, the turning point of the quadratic curve $y = x^2 + 8x - 10$, we need to calculate the vertex of the parabola.

The general formula for a parabola $y = ax^2 + bx + c$ gives the x-coordinate of the vertex as:

$x = -\frac{b}{2a}.$

Step 1: Identify coefficients

From the equation $y = x^2 + 8x - 10$:

• $a = 1$,
• $b = 8$,
• $c = -10$.

Step 2: Calculate x-coordinate of the vertex

$x = -\frac{8}{2(1)} = -4.$

Step 3: Calculate y-coordinate of the vertex

Substitute $x = -4$ into the original equation to find $y$: $y = (-4)^2 + 8(-4) - 10.$ $y = 16 - 32 - 10 = -26.$

Final Answer:

The turning point $P$ has coordinates: $P(-4, -26).$

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Related Questions:

1. How can we rewrite the equation $y = x^2 + 8x - 10$ in vertex form?
2. What is the axis of symmetry for the given quadratic equation?
3. How do we determine whether a turning point is a maximum or minimum?
4. Can we confirm the coordinates of $P$ by completing the square?
5. How does the graph shift if we modify the equation to $y = x^2 + 8x - 5$?

Tip:

To easily find the vertex of a quadratic equation, completing the square or using the vertex formula can save time.