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
[tex]\begin{array}{rcl}x + 3 = 0 & \qquad & x - 1 = 0\\x = \mathbf{-3} & \qquad & x = \mathbf{1}\\\end{array}[/tex]
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
[tex]\begin{array}{rr}\mathbf{x} & \mathbf{y} \\-2 & -2.8 \\0 & 3.0 \\2 & 4.8 \\4 & 2.6 \\6&-3.6\end{array}[/tex]
(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
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
[tex]\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}[/tex]
