Question

formula sheet matching. drag and drop the appropriate answer based on your formula sheet. linear formulas slope formula m = linear equations slope - intercept form: point - slope form: $y - y_1 = m(x - x_1)$ standard form: quadratic formulas quadratic equations standard form: vertex form: factored form: quadratic formula $x = $ drag & drop the answer $ax = by + c$ $y = mx + b$ $\frac{-b pm sqrt{b^2 - 4ac}}{2a}$ $\frac{-b pm sqrt{2b - 4ac}}{2a}$ $y = m + bx$ $\frac{-2a pm sqrt{b^2 - 4ac}}{b}$ $y = a(x + r)(x + s)$ $y = a(x - h)^2 + k$ $ax + by = c$ $y = a(x - r)(x - s)$ $\frac{y_2 - x_2}{y_1 - x_1}$ $y = a(x + h)^2 - k$

Explanation:

Step1: Slope Formula

The slope formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). So we match \(m=\) with \(\frac{y_2 - y_1}{x_2 - x_1}\).

Step2: Slope - Intercept Form of Linear Equation

The slope - intercept form of a linear equation is \(y=mx + b\), where \(m\) is the slope and \(b\) is the y - intercept. So we match "Slope - intercept Form:" with \(y = mx + b\).

Step3: Standard Form of Linear Equation

The standard form of a linear equation is \(ax+by = c\), where \(a\), \(b\) and \(c\) are real numbers and \(a\), \(b\) are not both zero. So we match "Standard Form:" (under Linear Equations) with \(ax + by=c\).

Step4: Standard Form of Quadratic Equation

The standard form of a quadratic equation is \(ax^{2}+bx + c=0\) (or we can also write it as \(ax^{2}+bx + c\) in the context of the formula sheet where we just need the polynomial form). But from the given options, if we consider the standard form as \(ax^{2}+bx + c = 0\) (the general quadratic polynomial form), but in the drag - and - drop options, if we assume the standard form of the quadratic equation (the polynomial form) is \(ax^{2}+bx + c\) (the non - factored, non - vertex form), but actually the standard form of a quadratic equation is \(ax^{2}+bx + c = 0\). However, from the given options, if we look at the quadratic equations' standard form, we can also think of it as \(ax^{2}+bx + c\) (the polynomial form). But maybe a better way: The standard form of a quadratic equation is \(ax^{2}+bx + c = 0\), but in the drag - and - drop options, if we have to choose from the given, maybe the standard form of the quadratic equation (the equation form) is \(ax^{2}+bx + c=0\), but in the options, we can also consider that the standard form of the quadratic function (if we are talking about the function) is \(y = ax^{2}+bx + c\). But in the given options, the closest for standard form of quadratic equation (the equation) is not directly given, but maybe the standard form of the quadratic function (as a polynomial) is \(ax^{2}+bx + c\). However, let's move to vertex form.

Step5: Vertex Form of Quadratic Equation

The vertex form of a quadratic function is \(y=a(x - h)^{2}+k\), where \((h,k)\) is the vertex of the parabola. So we match "Vertex Form:" with \(y=a(x - h)^{2}+k\).

Step6: Factored Form of Quadratic Equation

The factored form of a quadratic equation (or function) is \(y=a(x - r)(x - s)\), where \(r\) and \(s\) are the roots of the quadratic equation. So we match "Factored Form:" with \(y=a(x - r)(x - s)\).

Step7: Quadratic Formula

The quadratic formula for solving \(ax^{2}+bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). So we match \(x=\) with \(\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Answer:

• Slope Formula: \(m=\boldsymbol{\frac{y_2 - y_1}{x_2 - x_1}}\)
• Slope - intercept Form: \(y=\boldsymbol{mx + b}\)
• Standard Form (Linear): \(ax + by=\boldsymbol{c}\)
• Standard Form (Quadratic): \(ax^{2}+bx + c\) (or the equation form \(ax^{2}+bx + c = 0\), but from the options, if we consider the function form, and if the option for standard form of quadratic is \(ax^{2}+bx + c\))
• Vertex Form: \(y=\boldsymbol{a(x - h)^{2}+k}\)
• Factored Form: \(y=\boldsymbol{a(x - r)(x - s)}\)
• Quadratic Formula: \(x=\boldsymbol{\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}}\)

(Note: The above is the matching of each formula with the appropriate drag - and - drop option. The final answer is the set of correct matches as described above.)