Consider the quadratic f(x)=x^2-x-30Determine the following ( enter all numerical answers as integers,fraction or decimals$The smallest (leftmost) x-intercepts is x=The largest (rightmost)x-intercepts is x=The y-intercept is y=The vertex is The line of symmetry has the equation

Consider the quadratic fxx2x30Determine the following enter all numerical answers as integersfraction or decimalsThe smallest leftmost xintercepts is xThe large class=

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ANSWER

Smallest x-intercept: x = -5

Largest x-intercept: x = 6

y-intercept: y = -30

The vertex is (1/2, -121/4)

Line of symmetry x = 1/2

EXPLANATION

Given:

[tex]f(x)\text{ = x}^2\text{ - x - 30}[/tex]

Desired Results:

1. Smallest x-intercept: x =

2. Largest x-intercept: x =

3. y-intercept: y =

4. The vertex is

5. Equation of Line of symmetry

1. Determine the x-intercepts by equating f(x) to zero (0).

[tex]\begin{gathered} 0\text{ = x}^2-x-30 \\ x^2-6x+5x-30\text{ = 0} \\ x(x-6)+5(x-6)=0 \\ (x-6)(x+5)=0 \\ x-6=0,\text{ x+5=0} \\ x\text{ = 6, x = -5} \end{gathered}[/tex]

The smallest and largest x-intercepts are -5 and 6 respectively.

2. Determine the y-intercept by equating x to 0

[tex]\begin{gathered} y\text{ = \lparen0\rparen}^2-0-30 \\ y\text{ = -30} \end{gathered}[/tex]

y-intercept is -30

3a. Determine the x-coordinate of the vertex using the formula

[tex]x\text{ = -}\frac{b}{2a}[/tex]

where:

a = 1

b = -1

Substitute the values

[tex]\begin{gathered} x\text{ = -}\frac{(-1)}{2(1)} \\ x\text{ = }\frac{1}{2} \end{gathered}[/tex]

3b. Determine the y-coordinate of the vertex by substituting x into the equation

[tex]\begin{gathered} y\text{ = \lparen}\frac{1}{2})^2-\frac{1}{2}-30 \\ y\text{ = }\frac{1}{4}-\frac{1}{2}-30 \\ Find\text{ LCM} \\ y\text{ = }\frac{1-2-120}{4} \\ y\text{ = -}\frac{121}{4} \end{gathered}[/tex]

4. Determine the line of symmetry

In standard form the line of symmetry of a quadratic function can be identified using the formula

[tex]\begin{gathered} x\text{ = -}\frac{b}{2a} \\ x\text{ = }\frac{1}{2} \end{gathered}[/tex]