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Ch. 3 - Polynomial and Rational Functions
Blitzer - College Algebra 8th Edition
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Tutti i libri di testoBlitzer 8th EditionCh. 3 - Polynomial and Rational FunctionsProblema 97
Capitolo 4, Problema 97

In Exercises 97–98, write the equation of each parabola in vertex form. Vertex: (-3,-4) The graph passes through the point (1,4).

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Recall that the vertex form of a parabola's equation is given by y = a(x - h)2 + k, where (h, k) is the vertex of the parabola.
Substitute the vertex coordinates (-3, -4) into the vertex form equation, replacing h with -3 and k with -4, so the equation becomes y = a(x + 3)2 - 4.
Use the point (1, 4) that lies on the parabola to find the value of a. Substitute x = 1 and y = 4 into the equation: 4 = a(1 + 3)2 - 4.
Simplify the expression inside the parentheses and the square: 4 = a(4)2 - 4 becomes 4 = a imes 16 - 4.
Solve the resulting equation for a by isolating a on one side: add 4 to both sides and then divide by 16 to find the value of a.

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Vertex Form of a Parabola

The vertex form of a parabola's equation is y = a(x - h)^2 + k, where (h, k) is the vertex. This form makes it easy to identify the vertex and understand the parabola's shape and position on the coordinate plane.
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Vertex Form

Using a Point to Find the Parameter 'a'

After substituting the vertex coordinates into the vertex form, use another point on the parabola to solve for 'a'. This parameter controls the parabola's width and direction (upward if positive, downward if negative).
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Finding Equations of Lines Given Two Points

Substitution and Solving Quadratic Equations

Substitute the known point's x and y values into the vertex form equation to create an equation in terms of 'a'. Then solve this equation algebraically to find the exact value of 'a', completing the parabola's equation.
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Solving Systems of Equations - Substitution
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