Let's solve the quadratic function $f(x) = 2x^2 - 10x - 14$ by finding the required points step by step:

A) Finding the Vertex:

The vertex of a parabola given by $f(x) = ax^2 + bx + c$ is found using the formula: $x_{\text{vertex}} = -\frac{b}{2a}$ For $f(x) = 2x^2 - 10x - 14$, we have:

• $a = 2$
• $b = -10$

Using the vertex formula: $x_{\text{vertex}} = -\frac{-10}{2(2)} = \frac{10}{4} = 2.5$

To find the $y$-coordinate of the vertex, plug $x = 2.5$ back into the equation: $f(2.5) = 2(2.5)^2 - 10(2.5) - 14 = 2(6.25) - 25 - 14 = 12.5 - 25 - 14 = -26.5$

Thus, the vertex is $(2.5, -26.5)$.

B) Vertical Intercept (y-intercept):

The vertical intercept is the point where $x = 0$. Substituting $x = 0$ into the equation: $f(0) = 2(0)^2 - 10(0) - 14 = -14$ Thus, the vertical intercept is $(0, -14)$.

C) Finding the x-intercepts:

To find the x-intercepts, solve $f(x) = 0$: $2x^2 - 10x - 14 = 0$ We can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For $a = 2$, $b = -10$, and $c = -14$: $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(2)(-14)}}{2(2)} = \frac{10 \pm \sqrt{100 + 112}}{4} = \frac{10 \pm \sqrt{212}}{4}$ $\sqrt{212} \approx 14.56$ Thus, the x-intercepts are: $x = \frac{10 + 14.56}{4} = \frac{24.56}{4} = 6.14$ $x = \frac{10 - 14.56}{4} = \frac{-4.56}{4} = -1.14$

So, the x-intercepts are approximately $(6.14, 0)$ and $(-1.14, 0)$.

Summary:

• A) Vertex: $(2.5, -26.5)$
• B) Vertical intercept: $(0, -14)$
• C) x-intercepts: $(6.14, 0), (-1.14, 0)$

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you derive the quadratic formula?
2. What is the significance of the vertex in a quadratic function?
3. Can a parabola have more than two x-intercepts?
4. How does the value of 'a' affect the shape of the parabola?
5. What is the axis of symmetry for this quadratic function?

Tip: The discriminant $b^2 - 4ac$ in the quadratic formula tells you the nature of the roots: if positive, there are two real roots; if zero, one real root; if negative, no real roots.