Table of Contents
Last modified on October 9th, 2024
To graph a parabola, we first need to know its equation, which in the standard form is written as y = ax2 + bx + c
here,
However, the form of the equation can vary depending on whether the vertex is at the origin (0, 0) or any other point, say (h, k).
To graph a parabola, we need the following:
To graph a parabola, we first check whether the given equation is in standard form. If it is in standard form, we will proceed to step 1. If not, then we will first rewrite it in the standard form before following the given steps.
Let us graph the parabola f(x) = 2x2 – 6x + 4 …..(i)
Here, the parabola is already in its standard form.
Comparing the given equation with the standard form y = ax2 + bx + c, we get
a = 2, b = -6, c = 4
Finding y-intercept:
Setting x = 0, we get
y = f(x) = 2(0)2 – 6(0) + 4 = 4
Thus, the y-intercept is (0, 4)
Finding x-intercept:
Setting y = 0, we get
2x2 – 6x + 4 = 0
⇒ ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
⇒ ${x=\dfrac{-\left( -6\right) \pm \sqrt{\left( -6\right) ^{2}-4\times 2\times 4}}{2\times 2}}$
⇒ x = 2 and x = 1
Thus, the x-intercepts are (2, 0) and (1, 0)
The x-coordinate is given by ${h=-\dfrac{b}{2a}}$
Substituting the values, we get
h = ${-\dfrac{-6}{2\times 2}}$
⇒ h = ${\dfrac{3}{2}}$
⇒ h = 1.5
Now, substituting h into the equation (i), we get
k = f(h) = 2(1.5)2 – 6(1.5) + 4 = -0.5
⇒ k = -0.5
Thus, the vertex is (h, k) = (1.5, -0.5)
Now, we will form a function table with two columns (or rows) for the x-coordinate and y-coordinate, containing at least five points on the graph, including the vertex.
Since the vertex represents the ‘turning point’ of the parabola, the coordinates of the vertex will be placed in the middle of the function table. Here, (h, k) = (1.5, -0.5)
Now, we find points on each side of the vertex by choosing x-coordinates that are less than 1.5 and greater than 1.5 and substituting them into the function f(x) to obtain the corresponding f(x) values.
By considering two x-values less than 1.5 and two x-values greater than 1.5, we get
| x | f(x) |
| 0 | 4 |
| 1 | 0 |
| 1.5 | -0.5 |
| 2 | 0 |
| 3 | 4 |
Here,
At x = 0, f(0) = 2(0)2 – 6(0) + 4 = 4
At x = 1, f(1) = 2(1)2 – 6(1) + 4 = 0
At x = 2, f(2) = 2(2)2 – 6(2) + 4 = 0
At x = 3, f(3) = 2(3)2 – 6(3) + 4 = 4
Graph the equation of the parabola y = 3x2
Given y = 3x2, which is in the standard form.
Comparing the given equation with the standard form y = ax2 + bx + c, we get
a = 3, b = 0, c = 0
Now, setting x = 0, we get
y = f(x) = 3(0)2 = 0
Thus, the y-intercept is (0, 0)
Setting y = 0, we get
3x2 = 0
⇒ x = 0
Thus, the x-intercepts are (0, 0)
Here, the x-coordinate of the vertex is ${h=-\dfrac{b}{2a}}$ = ${-\dfrac{0}{2\times 3}}$ = ${0}$
⇒ h = 0
By substituting h = 0 in the equation y = 3x2, we get
k = f(h) = 3(0)2 = 0
⇒ k = 0
Thus, the vertex is (h, k) = (0, 0)
Here, the function table is
| x | f(x) |
| -2 | 12 |
| -1 | 3 |
| 0 | 0 |
| 1 | 3 |
| 2 | 12 |
When plotting the x-values and their corresponding y-values, we get the required graph.
Graph the parabola f(x) = -x2 + 4x – 1
Given f(x) = -x2 + 4x – 1, which is in the standard form.
Comparing the given equation with the standard form y = ax2 + bx + c, we get
a = -1, b = 4, c = -1
Now, putting x = 0, we get
y = f(x) = -(0)2 + 4(0) – 1 = -1
Thus, the y-intercept is (0, -1)
Setting y = 0, we get
-x2 + 4x – 1 = 0
⇒ ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
⇒ ${x=\dfrac{-4\pm \sqrt{4^{2}-4\times \left( -1\right) \times \left( -1\right) }}{2\times \left( -1\right) }}$
⇒ x = ${2+\sqrt{3}}$ and x = ${2-\sqrt{3}}$
Thus, the x-intercepts are ${\left( 2+\sqrt{3},0\right)}$ and ${\left( 2-\sqrt{3},0\right)}$
Here, the x-coordinate of the vertex is ${h=-\dfrac{b}{2a}}$ = ${-\dfrac{4}{2\times \left( -1\right) }}$ = ${2}$
⇒ h = 2
By substituting h = 2 in the equation f(x) = -x2 + 4x – 1, we get
k = f(h) = -(2)2 + 4(2) – 1 = 3
⇒ k = 3
Thus, the vertex is (h, k) = (2, 3)
Here, the function table is
| x | f(x) |
| 0 | -1 |
| 1 | 2 |
| 2 | 3 |
| 3 | 2 |
| 4 | -1 |
When plotting the x-values and their corresponding y-values, we get the required graph.
Graph: ${y=\dfrac{1}{2}x^{2}+x-1}$
Given ${y=\dfrac{1}{2}x^{2}+x-1}$
⇒ y = x2 + 2x – 2
Comparing the given equation with the standard form y = ax2 + bx + c, we get
a = 1, b = 2, and c = -2
Now, putting x = 0, we get
y = f(x) = (0)2 + 2(0) – 2 = -2
Thus, the y-intercept is (0, -2)
Putting y = 0, we get
x2 + 2x – 2 = 0
⇒ ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
⇒ ${x=\dfrac{-2\pm \sqrt{2^{2}-4\times 1 \times \left( -2\right) }}{2\times 1}}$
⇒ x = ${-1+\sqrt{3}}$ and x = ${-1-\sqrt{3}}$
Thus, the x-intercepts are ${\left( -1+\sqrt{3},0\right)}$ and ${\left( -1-\sqrt{3},0\right)}$
Here, the x-coordinate of the vertex is ${h=-\dfrac{b}{2a}}$ = ${-\dfrac{1}{2\times \dfrac{1}{2}}}$ = ${-1}$
⇒ h = -1
By substituting h = -1 in the equation ${y=\dfrac{1}{2}x^{2}+x-1}$, we get
k = f(h) = ${\dfrac{1}{2}\left( -1\right) ^{2}+\left( -1\right) -1}$
⇒ k = -1.5
Thus, the vertex is (-1, -1.5)
Here, the function table is
| x | f(x) |
| -4 | 3 |
| -2 | -1 |
| -1 | -1.5 |
| 0 | -1 |
| 2 | 3 |
When plotting the x-values and their corresponding y-values, we get the required graph.
Graph the parabola f(x) = -2x2 + 4x + 1
Given f(x) = -2x2 + 4x + 1, which is in the standard form.
Comparing the given equation with the standard form y = ax2 + bx + c, we get
a = -2, b = 4, c = 1
Now, putting x = 0, we get
y = f(x) = -2(0)2 + 4(0) + 1 = 1
Thus, the y-intercept is (0, 1)
Putting y = 0, we get
-2x2 + 4x + 1 = 0
⇒ ${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
⇒ ${x=\dfrac{-4\pm \sqrt{4^{2}-4\times \left( -2\right) \times 1}}{2\times \left( -2\right) }}$
⇒ x = ${1+\dfrac{\sqrt{6}}{2}}$ and x = ${1-\dfrac{\sqrt{6}}{2}}$
Thus, the x-intercepts are ${\left( 1+\dfrac{\sqrt{6}}{2},0\right)}$ and ${\left( 1-\dfrac{\sqrt{6}}{2},0\right)}$
Here, the x-coordinate of the vertex is ${h=-\dfrac{b}{2a}}$ = ${-\dfrac{4}{2\times \left( -2\right) }}$ = ${1}$
⇒ h = 1
By substituting h = 1 in the equation f(x) = -2x2 + 4x + 1, we get
k = f(h) = -2(1)2 + 4(1) + 1 = 3
⇒ k = 3
Thus, the vertex is (h, k) = (1, 3)
Here, the function table is
| x | f(x) |
| -1 | -5 |
| 0 | 1 |
| 1 | 3 |
| 2 | 1 |
| 3 | -5 |
When plotting the x-values and their corresponding y-values, we get the required graph.
Graph the parabola with the equation y = x2 – 4x + 3. Find its vertex, axis of symmetry and intercepts.
Given y = x2 – 4x + 3, which is in the standard form.
Comparing the given equation with the standard form y = ax2 + bx + c, we get
a = 1, b = -4, c = 3
Vertex
The x-coordinate of the vertex is ${h=-\dfrac{b}{2a}}$ = ${-\dfrac{-4}{2\times 1}}$ = ${2}$
⇒ h = 2
By substituting h = 2 in the equation y = x2 – 4x + 3, we get
k = (2)2 – 4(2) + 3 = -1
⇒ k = 15
Thus, the vertex is (h, k) = (2, -1)
Axis of Symmetry
The axis of symmetry is x = 2, which is the vertical line through the vertex.
y-intercept
Putting x = 0 in the equation y = x2 – 4x + 3,
y = (0)2 – 4(0) + 3 = 3
Thus, the y-intercept is (0, 3)
x-intercept
Putting y = 0 in the equation y = x2 – 4x + 3,
0 = x2 – 4x + 3
⇒ (x – 1)(x – 3) = 0
⇒ x = 1 and x = 3
Thus, the x-intercepts are (1, 0) and (3, 0)
Now, plotting the vertex, the axis of symmetry, and the intercepts on a graph, we get the required parabola.
Problem – Graphing a parabola in VERTEX FORM
Graph: y = -2(x – 1)2 + 4
Given y = -2(x – 1)2 + 4 …..(i)
As we know, the equation of the vertex form of a parabola is y = a(x – h)2 + k …..(ii)
Comparing the equations (i) and (ii), we get
a = -2, h = 1, and k = 4
Putting x = 0, we get
y = -2(0 – 1)2 + 4 = 2
Thus, the y-intercept is (0, 2)
Now, putting y = 0, we get
-2(x – 1)2 + 4 = 0
⇒ (x – 1)2 = 2
⇒ x = ${1+\sqrt{2}}$ and x = ${1-\sqrt{2}}$
Thus, the x-intercepts are ${\left( 1+\sqrt{2},0\right)}$ and ${\left( 1-\sqrt{2},0\right)}$
Also, the vertex is (1, 4)
Since a = -2 (negative), the parabola opens downwards.
Thus, the function table is
| x | f(x) |
| -1 | -4 |
| 0 | 2 |
| 1 | 4 |
| 2 | 2 |
| 3 | -4 |
When plotting the x-values and their corresponding y-values, we get the required graph.
Which of the following graphs is a parabola?
Here, option a) is a U-shaped curve. Thus, it is a parabola.
Last modified on October 9th, 2024