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Math  /  Algebra

QuestionExpress each quadratic function in standard form. State the yy-intercept of each. a) f(x)=−3(x−1)2+6f(x)=-3(x-1)^{2}+6 b) f(x)=4(x−3)(x+7)f(x)=4(x-3)(x+7)

Studdy Solution

STEP 1

What is this asking? We need to rewrite these quadratic functions in a different way, the *standard form*, and then find where they cross the *y*-axis. Watch out! Don't mix up the different forms of a quadratic function!
Remember, standard form looks like ax2+bx+cax^2 + bx + c.

STEP 2

1. Expand f(x)=−3(x−1)2+6f(x) = -3(x-1)^2 + 6
2. Find the *y*-intercept of f(x)=−3(x−1)2+6f(x) = -3(x-1)^2 + 6
3. Expand f(x)=4(x−3)(x+7)f(x) = 4(x-3)(x+7)
4. Find the *y*-intercept of f(x)=4(x−3)(x+7)f(x) = 4(x-3)(x+7)

STEP 3

First, let's **expand** that (x−1)2(x-1)^2 bit.
Remember, squaring something means multiplying it by itself!
So, (x−1)2=(x−1)⋅(x−1)(x-1)^2 = (x-1) \cdot (x-1).
Using the FOIL method (First, Outer, Inner, Last), we get x⋅x−x⋅1−1⋅x+1⋅1=x2−2x+1x \cdot x - x \cdot 1 - 1 \cdot x + 1 \cdot 1 = x^2 - 2x + 1.

STEP 4

Now, we **multiply** our result by −3-3.
So, −3⋅(x2−2x+1)=−3x2+6x−3-3 \cdot (x^2 - 2x + 1) = -3x^2 + 6x - 3.
Don't forget to distribute that −3-3 to every term inside the parentheses!

STEP 5

Almost there!
We just need to **add** 66 to our result.
So, −3x2+6x−3+6=−3x2+6x+3-3x^2 + 6x - 3 + 6 = -3x^2 + 6x + 3.
This is our **standard form**!

STEP 6

To find the *y*-intercept, we need to see where the graph crosses the *y*-axis.
This happens when x=0x = 0.
Let's **substitute** 00 for xx in our standard form equation: f(0)=−3(0)2+6(0)+3f(0) = -3(0)^2 + 6(0) + 3.

STEP 7

This simplifies to f(0)=3f(0) = 3.
So, our *y*-intercept is (0,3)(0, 3)!

STEP 8

Let's **expand** (x−3)(x+7)(x-3)(x+7) using FOIL again: x⋅x+x⋅7−3⋅x−3⋅7=x2+7x−3x−21=x2+4x−21x \cdot x + x \cdot 7 - 3 \cdot x - 3 \cdot 7 = x^2 + 7x - 3x - 21 = x^2 + 4x - 21.

STEP 9

Now, we **multiply** our result by 44: 4⋅(x2+4x−21)=4x2+16x−844 \cdot (x^2 + 4x - 21) = 4x^2 + 16x - 84.
This is the **standard form** for our second function!

STEP 10

Just like before, we **set** x=0x = 0 to find the *y*-intercept: f(0)=4(0)2+16(0)−84f(0) = 4(0)^2 + 16(0) - 84.

STEP 11

This simplifies to f(0)=−84f(0) = -84.
So, our *y*-intercept is (0,−84)(0, -84).

STEP 12

For f(x)=−3(x−1)2+6f(x) = -3(x-1)^2 + 6, the standard form is f(x)=−3x2+6x+3f(x) = -3x^2 + 6x + 3 and the *y*-intercept is (0,3)(0, 3).
For f(x)=4(x−3)(x+7)f(x) = 4(x-3)(x+7), the standard form is f(x)=4x2+16x−84f(x) = 4x^2 + 16x - 84 and the *y*-intercept is (0,−84)(0, -84).

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