Exercise 54 Page 391

There are certain points through which a graph passes that are more interesting than others.

• y-intercept
• x-intercept(s)
• vertex

The y-intercept we can find by determining the function's constant. Examining the function, we notice that it has a constant of 2. y=x^2+3x+ 2 ← constant This means the function intercepts the y-axis at y=2. To find the x-intercept(s) we want to rewrite the right-hand side in factored form. To do that, we can use a generic rectangle and a diamond problem. We know that x^2 and 2 goes into the lower left and upper right corners of the generic rectangle, respectively.

To fill in the remaining two corners, we need two x-terms that have a sum of 3x and a product of 2x^2.

Notice that both the product and the sum are positive. This means both factors must be positive. |c|c|c|c|c| [-1em] Product & ax(bx) & ax+bx & Sum & 3x? [0.2em] [-0.9em] 2x^2 & x(2x) &x+2x& 3x & ✓ [0.3em] When one term is x and the other is 2x, we have a product of 2x^2 and a sum of 3x. Now we can complete the diamond and generic rectangle.

To factor the right-hand side we add each side of the generic rectangle and multiply the sums. y=x^2+3x+2 ⇓ y=(x+1)(x+2) Having factored the right-hand side, we can find the x-intercepts by setting y equal to 0 and solving for x with the Zero Product Property.

y=(x+1)(x+2)

Substitute

y= 0

0=(x+1)(x+2)

▼

Solve for x

RearrangeEqn

Rearrange equation

(x+1)(x+2)=0

ZeroProdProp

Use the Zero Product Property

lcx+1=0 & (I) x+2=0 & (II)

SubEqn

(I): LHS-1=RHS-1

lx=-1 x+2=0

SubEqn

(II): LHS-2=RHS-2

lx_1=-1 x_2=-2

The graph has x-intercepts at x=-1 and x=-2. Finally, we want to identify the vertex. This can be done by rewriting the function in vertex form. Vertex Form:& y=(x- a)^2+ b Vertex:& ( a, b) If we rewrite the function to match this form exactly, we can identify the coordinates of the vertex. To complete the square we have to add the square of half the coefficient of x to both sides of the equation.

y=x^2+3x+2

AddEqn

LHS+(3/2)^2=RHS+(3/2)^2

y+(3/2)^2=x^2+3x+2+(3/2)^2

▼

Simplify

CalcQuot

Calculate quotient

y+1.5^2=x^2+3x+2+1.5^2

SplitIntoFactors

Split into factors

y+1.5^2=x^2+2(x)(1.5)+2+1.5^2

CommutativePropAdd

Commutative Property of Addition

y+1.5^2=x^2+2(x)(1.5)+1.5^2+2

FacPosPerfectSquare

a^2+2ab+b^2=(a+b)^2

y+1.5^2=(x+1.5)^2+2

CalcPow

Calculate power

y+2.25=(x+1.5)^2+2

SubEqn

LHS-2.25=RHS-2.25

y=(x+1.5)^2-0.25

AddPar

Add parentheses

y=(x-(-1.5))^2+(-0.25)

Now we can identify the coordinates of the vertex. Function:& y=(x-( -1.5))^2+( -0.25) Vertex:& ( -1.5, -0.25) The coordinates are (- 1.5,-0.25). Having found the most interesting points, we have enough information to make a decent sketch of the parabola.

As we can see, the graph is a parabola with two x-intercepts and a minimum value at (-1.5,-0.25).