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To solve this problem, let’s break down each part of the function \( f(x) = 2x^2 + 14x - 2 \).

### Step-by-Step Solution

#### A) The Vertex
The vertex of a parabola \( f(x) = ax^2 + bx + c \) can be found using the formula:
\[
x = -\frac{b}{2a}
\]
where \( a = 2 \) and \( b = 14 \).

1. Calculate the \( x \)-coordinate of the vertex:
   \[
   x = -\frac{14}{2 \times 2} = -\frac{14}{4} = -3.5
   \]

2. Substitute \( x = -3.5 \) back into the function to find the \( y \)-coordinate:
   \[
   f(-3.5) = 2(-3.5)^2 + 14(-3.5) - 2
   \]
   \[
   = 2(12.25) - 49 - 2 = 24.5 - 49 - 2 = -26.5
   \]

So, the vertex is:
\[
\text{Vertex} = (-3.5, -26.5)
\]

#### B) The Axis of Symmetry
The axis of symmetry is the vertical line that passes through the \( x \)-coordinate of the vertex. Therefore:
\[
\text{Axis of Symmetry} = x = -3.5
\]

#### C) The Y-Intercept
The \( y \)-intercept occurs when \( x = 0 \):
\[
f(0) = 2(0)^2 + 14(0) - 2 = -2
\]
So, the \( y \)-intercept is:
\[
\text{Y-intercept} = (0, -2)
\]

#### D) The Zeros of the Parabola (x-intercepts)
To find the zeros of the function, set \( f(x) = 0 \):
\[
2x^2 + 14x - 2 = 0
\]

This is a quadratic equation in standard form \( ax^2 + bx + c = 0 \), which we can solve using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

1. Substitute \( a = 2 \), \( b = 14 \), and \( c = -2 \):
   \[
   x = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}
   \]
   \[
   = \frac{-14 \pm \sqrt{196 + 16}}{4}
   \]
   \[
   = \frac{-14 \pm \sqrt{212}}{4}
   \]
   \[
   = \frac{-14 \pm 14.56}{4}
   \]

2. Calculate each solution:
   - For the \( + \) case:
     \[
     x = \frac{-14 + 14.56}{4} = \frac{0.56}{4} = 0.14
     \]
   - For the \( - \) case:
     \[
     x = \frac{-14 - 14.56}{4} = \frac{-28.56}{4} = -7.14
     \]

So, the zeros are:
\[
\text{Zeros} = (0.14, 0), (-7.14, 0)
\]

### Summary of Answers
- **A)** The vertex is \( (-3.5, -26.5) \).
- **B)** The axis of symmetry is \( x = -3.5 \).
- **C)** The \( y \)-intercept is \( (0, -2) \).
- **D)** The zeros are \( (0.14, 0) \) and \( (-7.14, 0) \).

Would you like further explanation on any part of this solution?

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Here are some related questions to deepen understanding:
1. How would the vertex change if the \( a \)-value in the equation were negative?
2. What does the \( y \)-intercept represent in the context of this parabola?
3. How do the zeros of the parabola relate to its graph on the coordinate plane?
4. What role does the axis of symmetry play in determining the shape of the parabola?
5. How can the quadratic formula be applied to other types of functions?

**Tip:** Always double-check calculations when using the quadratic formula, especially under the square root, to avoid common errors.

Consider the parabola given by the equation f(x) = 2x^2 + 14x - 2. Find the vertex, axis of symmetry, y-intercept, and zeros of the parabola.

Learn how to find the vertex, axis of symmetry, y-intercept, and zeros of the quadratic function f(x) = 2x^2 + 14x - 2. This problem covers algebraic concepts in quadratic functions, including vertex and quadratic formula applications. Suitable for Grades 9-10.

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