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The function \( h \) is defined as \( h(x)=5 x^{2}-3 \)

Ask by Boone Clark. in the United States
Jan 24,2025

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Answer

The function \( h(x) = 5x^2 - 3 \) is a quadratic function that opens upwards with its vertex at \( (0, -3) \) and y-intercept at \( -3 \).

Solution

Let's analyze the function \( h(x) = 5x^2 - 3 \). 1. **Identify the type of function**: This is a quadratic function because it can be expressed in the standard form \( ax^2 + bx + c \), where \( a = 5 \), \( b = 0 \), and \( c = -3 \). 2. **Determine the features of the function**: - **Vertex**: The vertex of a quadratic function in the form \( ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). - **Axis of symmetry**: The axis of symmetry is the vertical line that passes through the vertex, given by \( x = -\frac{b}{2a} \). - **Direction of opening**: Since \( a > 0 \), the parabola opens upwards. - **Y-intercept**: The y-intercept occurs when \( x = 0 \), which can be found by evaluating \( h(0) \). Let's calculate the vertex, axis of symmetry, and y-intercept. ### Step 1: Calculate the vertex Using the formula for the x-coordinate of the vertex: \[ x = -\frac{b}{2a} = -\frac{0}{2 \cdot 5} = 0 \] Now, substitute \( x = 0 \) back into the function to find the y-coordinate of the vertex: \[ h(0) = 5(0)^2 - 3 = -3 \] Thus, the vertex is at the point \( (0, -3) \). ### Step 2: Axis of symmetry The axis of symmetry is given by the line: \[ x = 0 \] ### Step 3: Y-intercept The y-intercept is: \[ h(0) = -3 \] ### Summary of features: - **Vertex**: \( (0, -3) \) - **Axis of symmetry**: \( x = 0 \) - **Direction of opening**: Upwards - **Y-intercept**: \( -3 \) If you need further analysis or specific calculations related to this function, please let me know!

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The Deep Dive

Did you know that quadratic functions like \( h(x) = 5x^2 - 3 \) have a parabolic shape, which is quite fascinating? This particular function opens upwards because the coefficient of \( x^2 \) is positive. The vertex of the parabola represents the minimum point, which in this case occurs at \( (0, -3) \). This means, regardless of how large or small the values of \( x \) become, \( h(x) \) can never go below -3! Now, if you're looking to apply this function in real-world contexts, think about any scenario involving acceleration, like the height of a projectile over time. The \( h(x) \) can represent the height of an object thrown upwards from a point -like a cannonball! Just plug in the time values (as \( x \)) and see how high your cannonball will go! Remember, the actual situation might require additional factors, like air resistance, but the fundamental principles hold true!

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