Determine the x-intercepts, axis of symmetry, turning point, and sketch the graph for the quadratic function y = 8 − 2x2.

Here's how to determine the features of the quadratic function $y = 8 - 2x^2$:

1.4.1 x-intercepts

To find the x-intercepts, set $y=0$ and solve for $x$.

$$0 = 8 - 2x^2$$

Add $2x^2$ to both sides:

$$2x^2 = 8$$

Divide by 2:

$$x^2 = 4$$

Take the square root of both sides:

$$x = \pm\sqrt{4}$$ $$x = \pm 2$$

The x-intercepts are at $(-2, 0)$ and $(2, 0)$.

$$\boxed{x = -2 \text{ and } x = 2}$$

1.4.2 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 the equation $y = 8 - 2x^2$, we can write it as $y = -2x^2 + 0x + 8$.
So, $a = -2$, $b = 0$, and $c = 8$.

$$x = -\frac{0}{2(-2)}$$ $$x = 0$$

The axis of symmetry is the y-axis.

$$\boxed{x = 0}$$

1.4.3 Turning point

The x-coordinate of the turning point is the axis of symmetry, which is $x=0$.
Substitute $x=0$ into the equation to find the y-coordinate:

$$y = 8 - 2(0)^2$$ $$y = 8 - 0$$ $$y = 8$$

The turning point is $(0, 8)$.

$$\boxed{(0, 8)}$$

1.4.4 Sketch the graph

To sketch the graph of $y = 8 - 2x^2$:

• It is a parabola opening downwards because the coefficient of $x^2$ (which is $-2$) is negative.
• The x-intercepts are at $(-2, 0)$ and $(2, 0)$.
• The y-intercept is found by setting $x=0$, which gives $y = 8 - 2(0)^2 = 8$. So, the y-intercept is $(0, 8)$. This is also the turning point.
• Plot these three points and draw a smooth parabolic curve through them, opening downwards.