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dmitriy555 [2]
2 years ago
8

Graph this parabola Show all steps please!!!

Mathematics
2 answers:
Margaret [11]2 years ago
5 0

In this case, the equation is in vertex form. To graph the parabola, we need to determine the y-intercept, x-intercept(s), and the vertex.

<h3><u>Vertex of Parabola:</u></h3>

Vertex form: y = a(x + h)² - k

  • ⇒ [y = 3(x + 2)² - 1] and [y = a(x - h)² - k]
  • ⇒ Vertex: (h, k) ⇒ (-2, -1)
<h3><u>X-intercept(s) of Parabola:</u></h3>

Assume "y" as 0.

\implies 0 = 3(x + 2)^{2}  - 1

\implies 1 = 3(x + 2)^{2}

<u>Take square root both sides and simplify:</u>

\implies\sqrt{1} = \sqrt{3(x + 2)^{2}}

\implies1 = \sqrt{3 \times (x + 2) \times (x + 2)}

\implies1 = (x + 2) \times \sqrt{3

<u>Divide √3 both sides:</u>

\implies \±\huge\text{(}\dfrac{1 }{\sqrt{3}} \huge\text{)} = (x + 2)    

\implies \±\huge\text{(}\dfrac{\sqrt{1}  }{\sqrt{3}} \huge\text{)} = (x + 2)    

<u>Multiply √3 to the numerator and the denominator:</u>

\implies \±\huge\text{(}\dfrac{\sqrt{1} \times \sqrt{3}  }{\sqrt{3\times \sqrt{3}}}  \huge\text{)} = (x + 2)

\implies \±\huge\text{(}\dfrac{\sqrt{3}  }{3}}  \huge\text{)} = (x + 2)

\implies \±\huge\text{(}\dfrac{\sqrt{3}  }{3}}  \huge\text{)} - 2 =x

\implies x = \underline{-2.6...} \ \text{and} \ \underline{-1.4...} \ \ \ \ (\text{Nearest tenth})

<h3><u>Y-intercept of Parabola:</u></h3>

Assume "x" as 0.

  • ⇒ y = 3(0 + 2)² - 1
  • ⇒ y = 3(2)² - 1
  • ⇒ y = 3(4) - 1
  • ⇒ y = 12 - 1
  • ⇒ y = 11

y-intercept = 11 ⇒ (0, 11)

<h3><u>Determining the direction of the parabola:</u></h3>

Since the first cooeficient is positive (+3), the direction of the parabola will be upwards.

<h3><u>Determining the axis of symmetry line:</u></h3>

Axis of symmetry line: x-coordinate of vertex

Axis of symmetry line: x = -2

<u>Plot the following on the graph:</u>

  • y-intercept ---> 11 -----> (0, 11)
  • x-intercept(s) -----> -1.4 and -2.6 -------> (-1.4, 0) and (-2.6, 0)
  • Vertex -------> (-2, -1)
  • Axis of symmetry ----> x = -2

Refer to graph attached.

ad-work [718]2 years ago
4 0

Answer:

<h3><u>Vertex</u></h3>

Vertex form of a quadratic equation:

y=a(x-h)^2+k\quad \textsf{where }(h,k)\:\textsf{is the vertex}

Given equation:  y=3(x+2)^2-1

Therefore, the vertex is (-2, -1)

<h3><u>x-intercepts (zeros)</u></h3>

The x-intercepts are when y=0:

\implies 3(x+2)^2-1=0

\implies 3(x+2)^2=1

\implies (x+2)^2=\dfrac13

\implies x+2=\pm\sqrt{\dfrac13}

\implies x+2=\pm\dfrac{\sqrt{3}}{3}

\implies x=-2\pm\dfrac{\sqrt{3}}{3}

\implies x=\dfrac{-6\pm\sqrt{3}}{3}

Therefore, the x-intercepts are at -2.6 and -1.4 (nearest tenth)

<h3><u>y-intercept</u></h3>

The y-intercept is when x=0:

\implies 3(0+2)^2-1=11

So the y-intercept is at (0, 11)

<h3><u>Plot the graph</u></h3>
  • As the leading coefficient is positive, the parabola will open upwards.
  • Plot the vertex at (-2, -1)
  • Plot the x-intercepts (-2.6, 0) and (-1.4, 0)
  • Plot the y-intercept (0, 11)

The axis of symmetry is the x-value of the vertex.  Therefore, the parabola is symmetrical about x = -2

**You don't need to plot the axis of symmetry - I have added it so that it is easier to draw the curve**

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