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A quadratic function is a type of polynomial function that has the general form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Quadratic functions are notable because they have a distinctive U-shaped graph called a parabola. Parabolas can open upwards or downwards depending on the sign of the leading coefficient \( a \). In our example, the quadratic function is given as \( f(x) = 3(x-3)^2 + 5 \). This form is known as the vertex form, where \( x - h \) is squared, and \( k \) is added. The vertex form makes it easy to identify the vertex of the parabola: the point \((h, k)\). This function's vertex is \((3, 5)\). Understanding the vertex form is crucial when you evaluate specific function values as you substitute different \( x \) values into the equation.

Algebraic expressions are combinations of numbers, variables, and operations (like addition, subtraction, multiplication, and division). The expression in our example is \( f(x) = 3(x - 3)^2 + 5 \).
An algebraic expression can represent a specific value for certain values of its variables. With our function, when you replace \( x \) with a number, you simplify the expression to find \( f(x) \).
The part of our expression \((x - 3)^2\) means you perform the operation \(x - 3\) first, then square the result. Afterward, you multiply by 3 and add 5, following the order of operations. Simplifying algebraic expressions involves systematically performing these operations to arrive at a number. This requires careful calculation and an understanding of how numbers and variables interact through mathematical operations.

The substitution method involves replacing a variable with a given number to evaluate an algebraic expression. It is a straightforward tactic used across various mathematical problems, including evaluating functions. In our example, to find \( f(3) \), \( f(-1) \), and \( f(0) \), you replace \( x \) in the expression \( f(x) = 3(x-3)^2 + 5 \) with these values one at a time.

• For \( f(3) \), substitute \( x = 3 \), resulting in \( 3(3-3)^2 + 5 \), which simplifies to 5.
• For \( f(-1) \), substitute \( x = -1 \), resulting in \( 3(-1-3)^2 + 5 \), which simplifies to 52.
• For \( f(0) \), substitute \( x = 0 \), resulting in \( 3(0-3)^2 + 5 \), which simplifies to 32.

Substitution merely requires attention to detail and patience, ensuring each arithmetic step is conducted in the correct order. This helps convert the algebraic function into easily calculable arithmetic expressions that provide the desired outputs.