In the image, you're working on a quadratic function homework, specifically Question 10, which asks you to sketch the graph of the quadratic function:

$f(x) = -2x^2 + 15$

Let’s break this down and solve it step by step.

1. Identify the equation type:

This is a quadratic function in the standard form $f(x) = ax^2 + bx + c$. For your function:

• $a = -2$ (which means the parabola opens downward),
• $b = 0$ (there is no linear term),
• $c = 15$ (this is the y-intercept).

2. Find the vertex:

The vertex form of a quadratic equation is $f(x) = a(x - h)^2 + k$, where $(h, k)$ is the vertex. Since there is no $b$ term, the vertex is located at $x = 0$.

Substitute $x = 0$ into the equation to find $y$ (the vertex's y-coordinate):

$f(0) = -2(0)^2 + 15 = 15$

So, the vertex is at $(0, 15)$.

3. Find the axis of symmetry:

The axis of symmetry is the vertical line that passes through the vertex. Since the vertex is at $x = 0$, the axis of symmetry is: $x = 0$

4. Find the intercepts:

• Y-intercept: This is the point where the graph crosses the y-axis ($x = 0$). We already know that $f(0) = 15$, so the y-intercept is $(0, 15)$.

• X-intercepts: To find the x-intercepts, set $f(x) = 0$ and solve for $x$:

$0 = -2x^2 + 15$ $2x^2 = 15$ $x^2 = \frac{15}{2}$ $x = \pm \sqrt{\frac{15}{2}} \approx \pm 2.74$

So the x-intercepts are approximately $x = 2.74$ and $x = -2.74$.

5. Sketch the graph:

• Plot the vertex at $(0, 15)$.
• Plot the x-intercepts at $(2.74, 0)$ and $(-2.74, 0)$.
• Since $a = -2$, the parabola opens downward.

You can now use this information to complete the graph in your homework using the graphing tool.

Would you like further details or have any questions?

5 Relative Questions:

1. How can you find the vertex of a quadratic function if the equation includes a linear term ($b \neq 0$)?
2. What is the significance of the sign of $a$ in determining the direction of the parabola?
3. How do you find the x-intercepts of a quadratic function when factoring isn't possible?
4. Why is the axis of symmetry always equal to $x = -\frac{b}{2a}$ for any quadratic function?
5. How does changing the value of $c$ in the equation affect the graph's position?

Tip:

Always check the discriminant $b^2 - 4ac$ when finding the x-intercepts to determine if real solutions (x-intercepts) exist!