Month: October 2018

Week 8 – Precalc 11

ximena

This week in Precalculus 11, we learned about quadratic functions and how to graph them. Quadratic functions are parabolas and they always have 2 variables (usually x and y), a degree of 2, a vertex, a line of symmetry, a domain, and a range. Both an x-intercept and a y-intercept are not mandatory but there usually at least one of them.

The Vertex is the highest or lowest point on the graph. If the parabola opens up, then the vertex is a minimum. If the parabola opens down, then the vertex is a maximum. The line of symmetry is the point where the parabola can be split in half evenly. The x-intercept is when the parabola touches the x-axis and the y-intercept is when the parabola touches the y-axis. The domain is the restrictions for the x-axis and range is the restrictions for the y-axis.

An example of a quadratic function:

y = x2+ x – 6

Since A is positive (1), we know that the parabola opens up and that the vertex is a minimum. We also know that C (-6) is the y-intercept.

To find the x-intercept roots, we have to make y = 0.

0 = x2+ x – 6

It’s now a quadratic equation, not a function, so we can solve for roots.

x2+ x – 6 = 0

(x + 3)(x – 2)

x = -3, 2

x-intercept roots: (-3, 0) and (2, 0)

Week 7 – Precalc 11

ximena

This week in pre-calculus 11, we learned about the discriminant. The discriminant is the radicand in the quadratic formula, b2– 4ac.

The discriminant is used to indicate whether the quadratic formula has 2 real roots or unequal roots, 1 real root or 2 equal roots, or no real roots.

x > 0: 2 real roots / 2 unequal roots.

x = 0: 1 real root / 2 equal roots.

x < 0: 0 real roots.

We use the quadratic equation to figure out which numbers are a, b, and c.

ax2 + bx + c = 0

Ex:

4x2 + 3x + 8 = 0

a = 4, b = 3, c = 8

b2– 4ac

32– 4(4)(8)

9 – 128

= -119

Seeing as the answer is negative, there are no real roots because -119 is less than 0.

Week 6 – Precalc 11

ximena

This week in Pre-calculus 11, I learned how to solve Perfect Square Trinomials.

To solve perfect square trinomials, the first and last terms in ax2 + bx + c have to be a perfect square.

Ex: x2 + 8x + 16

Then you have to factor by finding the number that multiplies twice into the last term (16), which is 4, and you have to make sure that the same number is the answer if the middle term, (a)x, is divided by 2. A different way to factor it is to square root the first and last term, and the answer for the last term square rooted should add together to equal the second/middle term.

4 x 4 = 16

8 ÷ 2 = 4

x2 + 8x + 16

= (x + 4)(x + 4)

= (x + 4)2

x = -4

To solve in the form of quadratic formula:

a = 1

b = 8

c = 16

 

*The discriminant equals 0, so we know that there can only be one real root/solution.

 

Week 5 – Precalc 11

ximena

This week in pre-calc 11, I learned how to factor polynomial expressions. C D P E U are the 5 steps to factoring a polynomial.

C = Common, D = Difference of squares, P = Pattern, E = Easy pattern, U = Ugly pattern

Common means finding a number or variable that can factor out from a binomial or a trinomial.

Ex. 64x + 16

= 16 (4x + 1)

 

Difference of squares is only used in binomials. The sign in between the expression has to be negative (-) and one number or variable has to be squared.

Ex. 25x2– 9

= (5x – 3)(5x + 3)

 

Pattern is to check if it’s a trinomial following the expression “(a)x2 + bx + c”.

– Easy patterns are when there is no coefficient in front of the x2. I then factor out to find the product of c and the sum of b with the same numbers.

Ex. x2 + 8x + 12

= (x + 6)(x+ 2)

– Ugly patterns are when there is a coefficient in front of the x2. My method is to multiply the coefficient into the term without a variable. It then becomes an easy pattern, then I factor it out. Once I get an answer with the easy pattern, I divide both sides by the coefficient of x2.

Ex. 2x2 + 3x – 2

= x2 + 3x – 4

= (x + 4)(x – 1)

= (x + \frac{4}{2})(x – \frac{1}{2})

= (x + 2)(2x – 1)