Solving Quadratic Systems

We want to solve the given system of inequalities by graphing. Note that both inequalities of the system are quadratic inequalities. ${y > x^{2} + 2 x y > x^{2} - 1 (I) (II)$ Let's graph each of them, one at a time.

Inequality (I)

We can write the boundary curve for Inequality (I) by replacing the greater than sign with an equals sign. Then, we can identify

a,

b,

and

c .

y = x^{2} + 2 x \Leftrightarrow y = 1 x^{2} + 2 x + 0

Knowing that

a = 1,

b = 2,

and

c = 0,

we can find the vertex. To do so, we will need to think of

y

as a function of

x,

y = f (x) .

Vertex of the parabola : (- \frac{b}{2 a}, f (- \frac{b}{2 a}))

Let's substitute the values of

a

and

b

in the formula for the

x

-coordinate of the vertex.

- \frac{b}{2 a}

SubstituteII

a = 1

,

b = 2

- \frac{2}{2 ( 1 )}

MultiplyMultiply

- \frac{2}{2}

CalcQuotCalculate quotient

- 1

The

x

-coordinate of the vertex is

- 1 .

Now, let's find the

y

-coordinate by substituting

- 1

for

x

into the quadratic equation for the boundary line.

y = x^{2} + 2 x

Substitute

x = - 1

y = (- 1)^{2} + 2 (- 1)

Simplify right-hand side

CalcPowCalculate power

y = 1 + 2 (- 1)

MultPosNeg

a (- b) = - a \cdot b

y = 1 - 2

SubTermSubtract term

y = - 1

The vertex is

(- 1, - 1) .

With this, we know that the axis of symmetry of the parabola is the vertical line

x = - 1 .

Next, let's find two more points on the curve, one on each side of the axis of symmetry.

$x$	$x^{2} + 2 x$	$y = x^{2} + 2 x$
$0$	$0^{2} + 2 (0)$	$0$
$- 2$	$(- 2)^{2} + 2 (- 2)$	$0$

The points $(0, 0)$ and $(- 2, 0)$ are on the parabola. Let's plot the points and connect them with a smooth curve.

Now that we have the boundary curve, we need to determine which region to shade. To do so, we will use

(- 1, 1)

as a test point. If the point satisfies the inequality, we will shade the region that contains the point. If not, we will shade the opposite region.

y > x^{2} + 2 x

SubstituteII

x = - 1

,

y = 1

1 > ? (- 1)^{2} + 2 (- 1)

Simplify right-hand side

CalcPowCalculate power

1 > ? 1 + 2 (- 1)

MultPosNeg

a (- b) = - a \cdot b

1 > ? 1 - 2

SubTermSubtract term

1 > - 1

Since the substitution produced a true statement, we will shade the region that contains the point $(- 1, 1) .$ Because we have a strict inequality, the boundary curve will be dashed.

Inequality (II)

Once again, we can write the boundary curve by replacing the greater than sign with an equals sign. $y = x^{2} - 1$ We can draw the second parabola following the same procedure as with the first.

Vertex	Axis of Symmetry	Two Points
$(0, - 1)$	$x = 0$ $(y$ -axis)	$(1, 0)$ and $(- 1, 0)$

Any point on the plane can be used to determine the region we should shade. Because this inequality is strict, the curve will be dashed.

Solution

The solution set is the overlapping region.