Question

HELP NOW PLZ 20 POINTS FOR BRAINLIEST Step 1: Write a quadratic equation in standard form choosing your own coefficients and constant! (Recall: Standard Form of a quadratic equation is Ax^2 + Bx + C ). You need to choose a value for A, B, and C that satisfies each of the following conditions. You can use a value of 0 for B for no more than 1 of the 5 questions! Step 2: Solve each of the equations you create. YOU MUST SHOW YOUR WORK TO RECEIVE CREDIT FOR THIS STEP! Write your Exact answer in simplest form AND if necessary, round your estimate to the nearest tenth. 1) Equation 1 should represent a parabola that opens up and has a negative y-intercept. Equation 1: ____________________________________________________________ What strategy are you using to solve this equation and why? ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ Show your work and solution for solving this equation: 2) Equation 2 should represent a parabola that opens down and has a negative y- intercept. Equation 2:___________________________________________________________________ What strategy are you using to solve this equation and why? ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ Show your work and solution for solving this equation: 3) Equation 3 should represent a parabola that is a vertical stretch of the parent function and has a y-intercept greater than 3 and opens down. Equation 3:___________________________________________________________________ What strategy are you using to solve this equation and why? ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ Show your work and solution for solving this equation: 4) Equation 4: You decide what type of parabola you would like to create by answer the following questions: Does your parabola open up or down? _______________ Is there a vertical stretch or vertical compression? _______ What is the y-intercept? __________ Equation 4: ____________________________________________________________________ What strategy are you using to solve this equation and why? ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ Show your work and solution for solving this equation: 5) Write a quadratic equation that can only be solved using the quadratic formula. Equation 4: _________________________________________________________________ Show your work for how you use the quadratic formula to solve this equation:

Answer 1

Answer:

Here's what I get  

Step-by-step explanation:

y = ax² + bx + c

1. Equation 1

I will use the equation

y = x² + 2x - 3

a = 1; b= 2; c = -3

a = 1, so the parabola opens up.

c < 0, so the y-intercept is negative.

Strategy:

I will factor and use the zero-product property because

3(-1) = -3 and 3 + (-1) = 2

(a) Find two numbers that multiply to give ac (13), and add to give b (2).

Factors of 2: +1,±2

Factors of 3: ±1, ±3

After some trial and error, you find the numbers are 3 and -1.

3 - 1 = 2 and 3(-1) = -3

(b) Rewrite the middle with those numbers

x² + 3x + (-1)x - 3

(c) Factor the first two and last two terms separately

x(x +3) + (-1)(x + 3)

(d) Take out the common factor

(x + 3)(x - 1)

(e) Apply the zero-product property

$\begin{array}{rcl}x + 3 = 0 & \qquad & x - 1 = 0\\x = \mathbf{-3} & \qquad & x = \mathbf{1}\\\end{array}$

2. Equation 2

y = -x² + 2x - 1

a = -1; b = 2; c = -1

a = -1, so the parabola opens down.

c < 0, so the y-intercept is negative.

Strategy:

I will solve this one by completing the square because it is less obvious how to factor a quadratic with a negative leading coefficient.

-x² + 2x - 1 = 0

(a) Divide each side by -1

x² - 2x + 1 = 0

(b) Move the constant term to the right-hand side

x² - 2x = -1

(c) Take half the coefficient of x, square it, and add it to each side

-2/2 = 1

   1² = 1

x² - 2x +1 = 0

(d) Factor each side as the square of a binomial

(x - 1)² = 0

(e) Take the square root of each side

x - 1 = 0

(f) Solve for x

x = 1

3. Equation 3

y = -½x² + 1.9x + 3

a = -½; b = 1.9; c = +3

a = -½, so the parabola opens down and is compressed vertically.

c = 2, so the y-intercept is at y = 3

Strategy:

We could I will solve this one graphically because the fractional coefficients look problematic.

(a) Make a table containing a few points

$\begin{array}{rr}\mathbf{x} & \mathbf{y} \\-2 & -2.8 \\0 & 3.0 \\2 & 4.8 \\4 & 2.6 \\6&-3.6\end{array}$

(b) Plot the points

(c) Join the points with a smooth curve

(d) Extend the curve to the edges of  the graph

The graph below shows your calculated points and the x-intercepts at  

x = -1.2 and x = 5

4. Equation 4

2x² - 8 = 0

a = 2; b = 0; c = -8

a = 2, so the parabola opens upward and is vertically stretched.

c = -8, so the y-intercept is at y = -8.

Strategy:

b = 0, so I will use the method of taking the square root of each side.

(a) Remove the common factor

2x² - 8 = 0

  x² - 4 = 0

(b) Move the constant term to the other side

x² = 4

(c) Take the square root of each side

x = ±2

5. Equation 5

y = x² + 3x + 1

This equation has no rational factors, so the formula will be useful.

a = 1; b = 1, c = 1

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} \\\\ & = & \dfrac{-3\pm\sqrt{3^2 - 4\times 1\times 1}}{2\times 1} \\\\ & = & \dfrac{-3\pm\sqrt{9 - 4}}{-2} \\\\ & = & \dfrac{-3\pm\sqrt{5}}{-2} \\\\x=\dfrac{-3 + \sqrt{5}}{-2} & &x=\dfrac{-3 - \sqrt{5}}{-2}\\\\x=\mathbf{\dfrac{3}{2} - \dfrac{\sqrt{5}}{2}}& & x = \mathbf{\dfrac{3}{2} + \dfrac{\sqrt{5}}{2}}\\\\\end {array}$