To solve the system of equations by graphing, we will draw the graph of the quadratic function and the linear function on the same coordinate grid. Let's start with the parabola.

Graphing the Parabola

To graph the parabola, we first need to identify

a,

b,

and

c .

y = x^{2} + 4 x - 4 \Leftrightarrow y = 1 x^{2} + 4 x + (- 4)

For this equation we have that

a = 1,

b = 4,

and

c = - 4 .

Now, we can find the vertex using its formula. To do this, we will need to think of

y

as a function of

x,

y = f (x) .

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

Let's find the

x -

coordinate of the vertex.

- \frac{b}{2 a}

SubstituteII

$a = 1$ , $b = 4$

- \frac{4}{2 ( 1 )}

MultByOne

$a \cdot 1 = a$

- \frac{4}{2}

CalcQuot

Calculate quotient

- 2

We use the

x -

coordinate of the vertex to find its

y -

coordinate by substituting it into the given equation.

y = x^{2} + 4 x - 4

Substitute

$x = - 2$

y = (- 2)^{2} + 4 (- 2) - 4

▼

Simplify right-hand side

NegBaseToPosBase

$(- a)^{2} = a^{2}$

y = 4 + 4 (- 2) - 4

MultPosNeg

$a (- b) = - a \cdot b$

y = 4 - 8 - 4

SubTerms

Subtract terms

y = - 8

The

y -

coordinate of the vertex is

- 8 .

Thus, the vertex is at the point

(- 2, - 8) .

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

x = - 2 .

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

$x$	$x^{2} + 4 x - 4$	$y = x^{2} + 4 x - 4$
$0$	$0^{2} + 4 (0) - 4$	$- 4$
$- 4$	$(- 4)^{2} + 4 (- 4) - 4$	$- 4$

Both $(0, - 4)$ and $(- 4, - 4)$ are on the graph. Let's form the parabola by connecting these points and the vertex with a smooth curve.

Graphing the Line

Let's now graph the linear function on the same coordinate plane. For a linear equation written in slope-intercept form, we can identify its slope

m

and

y -

intercept

b .

y = 2 x - 5 \Leftrightarrow y = 2 x + (- 5)

The slope of the line is

2

and the

y -

intercept is

- 5 .

Finding the Solutions

Finally, let's try to identify the coordinates of the point of intersection of the parabola and the line.

It looks like the point of intersection occurs at $(- 1, - 7) .$

Checking the Answer

To check our answers, we will substitute the value of the point of intersection in both equations of the system. If it produces true statements, our solution is correct. Let's do it!

{y = x^{2} + 4 x - 4 y = 2 x - 5 (I) (II)

$(I), (II):$ $x = - 1$ , $y = - 7$

{- 7 = ? (- 1)^{2} + 4 (- 1) - 4 - 7 = ? 2 (- 1) - 5

▼

(I), (II):

Simplify right-hand side

NegBaseToPosBase

$(I):$ $(- a)^{2} = a^{2}$

{- 7 = ? 1 + 4 (- 1) - 4 - 7 = ? 2 (- 1) - 5

$(I), (II):$ $a (- b) = - a \cdot b$

{- 7 = ? 1 - 4 - 4 - 7 = ? - 2 - 5

$(I), (II):$ Subtract terms

{- 7 = - 7 ✓ - 7 = - 7 ✓

Equation (I) and Equation (II) both produced true statements. Therefore,

(- 1, - 7)

is a correct solution.

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