Converting Parabola Equations: Standard to Vertex Form
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Learn the step-by-step process of 'completing the square' to convert a quadratic equation from standard form (ax^2 + bx + c) to vertex form (a(x-h)^2 + k). This skill is crucial for quickly identifying a parabola's vertex.
Converting Parabola Equations: Standard to Vertex Form
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What is the primary method used to convert a quadratic equation from standard form to vertex form?
The method is called "completing the square."
What is the goal of completing the square for a quadratic \(ax^2+bx+c\)?
The goal is to rewrite the expression by creating a perfect square trinomial, \((x-h)^2\), which can then be factored. This reveals the vertex \((h, k)\).
Given an expression \(x^2 + Bx\), what term must be added to "complete the square"?
You must add \((\frac{B}{2})^2\). This creates the perfect square trinomial \(x^2 + Bx + (\frac{B}{2})^2\), which factors to \((x + \frac{B}{2})^2\).
Convert \(y = x^2 + 6x + 5\) to vertex form. Step 1: Isolate the x-terms.
Group the x-terms and move the constant: \(y = (x^2 + 6x) + 5\).
For \(y = (x^2 + 6x) + 5\), what number completes the square inside the parentheses, and how do you balance the equation?
To complete the square for \(x^2 + 6x\), you add \((\frac{6}{2})^2 = 3^2 = 9\). To keep the equation balanced, you must also subtract \(9\) outside the parentheses. \(y = (x^2 + 6x + 9) + 5 - 9\).
After completing the square, factor the trinomial and simplify the expression \(y = (x^2 + 6x + 9) + 5 - 9\).
The trinomial factors to \((x+3)^2\). The constants simplify to \(-4\). The final vertex form is \(y = (x+3)^2 - 4\).
What is the vertex of the parabola \(y = x^2 + 6x + 5\)?
By converting to vertex form \(y = (x+3)^2 - 4\), we can identify the vertex as \((-3, -4)\).
When converting \(y = ax^2 + bx + c\) to vertex form where \(a \neq 1\), what is the first step?
Factor the leading coefficient, \(a\), from only the first two terms: \(y = a(x^2 + \frac{b}{a}x) + c\).
Convert \(y = 2x^2 + 12x + 10\) to vertex form. Start by factoring \(a\).
Factor \(2\) from the first two terms: \(y = 2(x^2 + 6x) + 10\).
For \(y = 2(x^2 + 6x) + 10\), what do you add inside the parentheses to complete the square?
For the expression \(x^2+6x\), you add \((\frac{6}{2})^2 = 9\) inside the parentheses. So you have \(y = 2(x^2 + 6x + 9) + ...\)
When adding \(9\) inside the parentheses of \(y = 2(x^2 + 6x) + 10\), what value must be subtracted outside to balance the equation?
Because the \(9\) is inside parentheses being multiplied by \(2\), you have actually added \(2 \times 9 = 18\) to the equation. Therefore, you must subtract \(18\) outside the parentheses.
Finish converting \(y = 2x^2 + 12x + 10\) to vertex form after setting it up as \(y = 2(x^2 + 6x + 9) + 10 - 18\).
Factor the perfect square trinomial and combine the constants: \(y = 2(x+3)^2 - 8\).
Convert the equation \(y = -3x^2 + 6x - 1\) to vertex form.
