Question

Graph this parabola Show all steps please!!!

Answer 1

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.

Vertex of Parabola:

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)

X-intercept(s) of Parabola:

Assume "y" as 0.

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

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

Take square root both sides and simplify:

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

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

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

Divide √3 both sides:

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

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

Multiply √3 to the numerator and the denominator:

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

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

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

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

Y-intercept of Parabola:

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)

Determining the direction of the parabola:

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

Determining the axis of symmetry line:

Axis of symmetry line: x-coordinate of vertex

Axis of symmetry line: x = -2

Plot the following on the graph:

• 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.

Answer 2

Answer:

Vertex

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)

x-intercepts (zeros)

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)

y-intercept

The y-intercept is when $x=0$:

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

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

Plot the graph

• 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**