Math / Geometry

QuestionSolve $t$ nis pair of simultancous equation graphically, $y=x^{2}+4 x-5$ and $y=2 x-2$ . And also shade the region defined simultaneously by the inequalities $y \geq x^{2}+4 x-5$ and $y \leq 2 x-2$ on the same graph.
Unit 11.2 Graphs \& Functions ... Topici: Quadratic Equations \& Quadratic Gri

Studdy Solution

STEP 1

1. We need to solve the simultaneous equations $y = x^2 + 4x - 5$ and $y = 2x - 2$ graphically.
2. We need to find the points of intersection of the two curves.
3. We need to shade the region defined by the inequalities $y \geq x^2 + 4x - 5$ and $y \leq 2x - 2$ .
4. The solutions and shaded regions will be identified graphically on the same coordinate plane.

STEP 2

1. Graph the quadratic equation $y = x^2 + 4x - 5$ .
2. Graph the linear equation $y = 2x - 2$ .
3. Find the points of intersection of the two curves.
4. Identify and shade the region defined simultaneously by the inequalities $y \geq x^2 + 4x - 5$ and $y \leq 2x - 2$ .

STEP 3

Graph the quadratic equation $y = x^2 + 4x - 5$ by plotting several points. To find the vertex and the axis of symmetry, use the formula for the vertex $x = -\frac{b}{2a}$ for the quadratic equation $ax^2 + bx + c$ .
For $y = x^2 + 4x - 5$ , $a = 1$ , $b = 4$ , $c = -5$ . The vertex is at $x = -\frac{4}{2 \cdot 1} = -2$ . Calculate $y$ at $x = -2$ :
$y = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9$
Thus, the vertex is $(-2, -9)$ . Plot this point and other points by choosing values for $x$ and solving for $y$ .

STEP 4

Plot additional points for the quadratic equation to better define its shape. For example, calculate $y$ for $x = -3, -1, 0, 1, 2$ :
For $x = -3$ : $y = (-3)^2 + 4(-3) - 5 = 9 - 12 - 5 = -8$
For $x = -1$ : $y = (-1)^2 + 4(-1) - 5 = 1 - 4 - 5 = -8$
For $x = 0$ : $y = 0^2 + 4(0) - 5 = -5$
For $x = 1$ : $y = 1^2 + 4(1) - 5 = 1 + 4 - 5 = 0$
For $x = 2$ : $y = 2^2 + 4(2) - 5 = 4 + 8 - 5 = 7$
Plot these points on the graph and draw the parabola.

STEP 5

Graph the linear equation $y = 2x - 2$ by finding the y-intercept and slope. The y-intercept is at $(0, -2)$ , and the slope is $2$ .
Plot the point $(0, -2)$ . For another point, use $x = 1$ : $y = 2(1) - 2 = 0$
Thus, the point $(1, 0)$ is also on the line. Draw the line through these points.

STEP 6

Find the points of intersection by setting the equations equal to each other and solving for $x$ :
$x^2 + 4x - 5 = 2x - 2$
Rearrange the equation: $x^2 + 4x - 5 - 2x + 2 = 0$ $x^2 + 2x - 3 = 0$
Factor the quadratic: $(x + 3)(x - 1) = 0$
Solve for $x$ : $x = -3 \quad \text{or} \quad x = 1$
Substitute these $x$ values back into $y = 2x - 2$ to find the corresponding $y$ values: For $x = -3$ : $y = 2(-3) - 2 = -6 - 2 = -8$ So, the point is $(-3, -8)$ .
For $x = 1$ : $y = 2(1) - 2 = 0$ So, the point is $(1, 0)$ .
Plot these intersection points on the graph.

STEP 7

Shade the region defined by the inequalities $y \geq x^2 + 4x - 5$ and $y \leq 2x - 2$ . This is the region above the parabola and below the line.
First, identify the region above the parabola $y = x^2 + 4x - 5$ . This is the area where $y$ values are greater than or equal to those on the parabola.
Next, identify the region below the line $y = 2x - 2$ . This is the area where $y$ values are less than or equal to those on the line.
The shaded region is where these two areas overlap.
Solution: The solutions to the simultaneous equations are the points of intersection: $(-3, -8)$ and $(1, 0)$ .
The shaded region on the graph represents the area above the parabola $y = x^2 + 4x - 5$ and below the line $y = 2x - 2$ .

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