**Exploring graphs of quadratic equations :**

The graph of any quadratic equation will be a parabola.

The general form of a quadratic equation is

ax² + bx + c = 0

To graph any quadratic equation, we have to equate ax² + bx + c to y.

Then, we will have the general form of quadratic function.

That is,

y = ax² + bx + c

To explore the graphs of quadratic functions, we have to be aware of the following stuff.

(i) Open upward or open downward parabola

(ii) Vertex

(iii) x - intercepts

(x - intercepts are nothing but the points where the curve cuts x - axis)

(iv) y - intercept

(y - intercepts are nothing but the points where the curve cuts y - axis)

**(i) Open upward or downward :**

y = ax² + bx + c

If the leading coefficient or the sign of "a" is positive, the parabola is open upward and "a" is negative, the parabola is open downward.

**(ii) Vertex :**

To find the vertex of the parabola which is given by the quadratic function y = ax² + bx + c, we have to find the value "x" using the formula given below.

**x = -b / 2a**

And the vertex is

**[ y(-b/2a) , -b/2a ]**

**(iii) X - intercepts :**

To know the x - intercepts, we have to plug y = 0 in y = ax² + bx + c and find the values of "x".

**Note : **

If the values of "x" are imaginary, the curve will not intersect x - axis.

**(iii) Y - intercept :**

To know the y - intercept, we have to plug x = 0 in y = ax² + bx + c and find the value of "y".

Based on the results which we get from the above four points, we have to sketch parabola.

**Example 1 : **

Graph the quadratic equation that is given below.

x² + 5x + 6 = 0

**Solution : **

Let y = x² + 5x + 6

**Step 1 :**

In the given quadratic function, since the leading coefficient (x²) is positive, the parabola will be open upward.

**Step 2 :**

When we compare the given quadratic function with y = ax² + bx + c, we get

a = 1

b = 5

c = 6

"x" coordinate of the vertex = -b / 2a

"x" coordinate of the vertex = -5 / 2(1)

"x" coordinate of the vertex = -5 / 2

"x" coordinate of the vertex = -2.5

"y" coordinate of the vertex = y(-2.5)

"y" coordinate of the vertex = (-2.5)² + 5(-2.5) + 6

"y" coordinate of the vertex = 6.25 - 12.5 + 6

"y" coordinate of the vertex = -0.25

Vertex (-2.5, -0.25)

**Step 3 :**

To know x - intercept, we have to plug y = 0 in the given quadratic function.

Then, we have

0 = x² + 5x + 6 or x² + 5x + 6 = 0

(x + 2)(x + 3) = 0

x + 2 = 0 or x + 3 = 0

x = -2 or x = -3

Therefore, the parabola cuts x - axis at x = -2 and x = -3

**Step 4 : **

To know y - intercept, we have to plug x = 0 in the given quadratic function.

Then, we have

y = (0)² + 5(0) + 6

y = 6

Therefore, the parabola cuts y - axis at y = 6.

Now, let us sketch the parabola.

**Example 2 : **

Graph the quadratic equation that is given below.

-2x² + 5x - 7 = 0

**Solution : **

**Let y = -2x² + 5x - 7**

**Step 1 :**

In the given quadratic function, since the leading coefficient (x²) is negative, the parabola will be open downward.

**Step 2 :**

When we compare the given quadratic function with y = ax² + bx + c, we get

a = -2

b = 5

c = -7

"x" coordinate of the vertex = -b / 2a

"x" coordinate of the vertex = -5 / 2(-2)

"x" coordinate of the vertex = -5 / (-4)

"x" coordinate of the vertex = 1.25

"y" coordinate of the vertex = y(1.25)

"y" coordinate of the vertex = -2(1.25)² + 5(1.25) - 7

"y" coordinate of the vertex = -3.125 + 6.25 - 7

"y" coordinate of the vertex = -3.875

Vertex (1.25, -3.875)

**Step 3 :**

To know x - intercept, we have to plug y = 0 in the given quadratic function.

Then, we have

0 = -2x² + 5x -7 or -2x² + 5x - 7 = 0

The above quadratic equation can not be solved using factoring.

So let us try to solve the equation using quadratic formula as given below

Clearly, the two values of "x" are imaginary.

Therefore, the parabola does not cut x - axis.

**Step 4 : **

To know y - intercept, we have to plug x = 0 in the given quadratic function.

Then, we have

y = -2(0)² + 5(0) - 7

y = -7

Therefore, the parabola cuts y - axis at y = -7.

Now, let us sketch the parabola.

After having gone through the stuff given above, we hope that the students would have understood "Exploring graphs of quadratic equations".

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