Problem 42 - Intermediate Algebra

Question

If \(h(x)=-x^{2}-3\), find \(h(1), h(-1), h(-3)\), and \(h(5)\).

Step-by-Step Solution

Answer

\( h(1) = -4 \), \( h(-1) = -4 \), \( h(-3) = -12 \), \( h(5) = -28 \).

Step 1: Substitute into the Function

To find the value of the function at a particular point, substitute the given x-value into the function. Let's start with finding \( h(1) \).

Step 2: Calculate \( h(1) \)

Substitute 1 for \( x \) in the function: \[h(1) = -1^2 - 3 = -1 - 3 = -4.\]Thus, \( h(1) = -4 \).

Step 3: Calculate \( h(-1) \)

Substitute -1 for \( x \) in the function: \[h(-1) = -(-1)^2 - 3 = -1 - 3 = -4.\]Thus, \( h(-1) = -4 \).

Step 4: Calculate \( h(-3) \)

Substitute -3 for \( x \) in the function: \[h(-3) = -(-3)^2 - 3 = -9 - 3 = -12.\]Thus, \( h(-3) = -12 \).

Step 5: Calculate \( h(5) \)

Substitute 5 for \( x \) in the function: \[h(5) = -5^2 - 3 = -25 - 3 = -28.\]Thus, \( h(5) = -28 \).

Key Concepts

Quadratic Functions

A quadratic function is special because its highest power of the variable is squared. It's usually written in the form of \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a \) isn't zero. A key feature of quadratic functions is that they create a U-shaped curve called a parabola when graphed. Depending on the sign of \( a \), this parabola can either open upwards or downwards.

• If \( a > 0 \), the parabola opens upwards, resembling a 'U'.
• If \( a < 0 \), the parabola opens downwards, like an upside-down 'U'.

In the function from the exercise, \( h(x) = -x^2 - 3 \), it's apparent that \( a = -1 \), which means the parabola is opening downwards. Notice there are no \( b \) terms, simplifying our calculations because we focus only on the quadratic term \( x^2 \), and the constant \(-3\), which shifts the parabola downward by 3 units. This understanding helps establish a visual expectation when calculating function values.

Negative Exponents

Dealing with exponents can sometimes be tricky, but understanding their nature makes it easier. A negative exponent signifies that the base is on the wrong side of a fraction. To simplify, you take the reciprocal of the base and switch the negative exponent to positive. For example, \( x^{-n} = \frac{1}{x^n} \).

In the context of quadratic functions like \( h(x) = -x^2 - 3 \), even though we don't directly encounter negative exponents, it's important to know this concept. This understanding informs us on how exponents influence equations. Thankfully, the calculation remains straightforward since squaring always results in a non-negative value. It's pivotal to remember that negative signs within and outside the square can impact results, such as \((-3)^2\) being \(9\), and not \(-9\). This distinction is crucial when working through substitutions.

Substitution Method

The substitution method is a straightforward technique used in mathematics. It helps to determine the value of a function for specific values of \( x \) by directly replacing \( x \) in the equation with those values.

For this exercise, if we want to find \( h(1) \), we substitute \( 1 \) for \( x \) in the equation \( h(x) = -x^2 - 3 \). So the equation becomes \( -1^2 - 3 \) which simplifies down to \( -4 \). This method reliably calculates function values because it simplifies how we approach equations, focusing on arithmetic after substitution.

• Pick the correct value for substitution.
• Simplify according to the arithmetic operations.
• Verify your calculations for signs and order of operations.

This approach is especially useful for evaluating quadratic or any polynomials where direct plug-in, simplify, and solve yields results efficiently.