#### Using Intercept Form

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###### Monitoring Progress
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###### Exercises
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Exercises 10 To draw the graph of the given function, we will follow four steps.Identify and plot the x-intercepts. Find and graph the axis of symmetry. Find and plot the vertex. Draw the parabola through the vertex and the points where the x-intercepts occur.Let's go through these steps one at a time.Identify and Plot the x-intercepts Let's start by recalling the intercept form of a quadratic function. f(x)=a(x−p)(x−q)​ In this form, where a ≠ 0, the x-intercepts are p and q. Let's consider the given function. h(x)=-4(x−7)(x−3)​ We can see that a=-4, p=7, and q=3. Therefore, the x-intercepts occur at (7,0) and (3,0).Find and Graph the Axis of Symmetry The axis of symmetry is halfway between (p,0) and (q,0). Since we know that p=7 and q=3, the axis of symmetry of our parabola is halfway between (7,0) and (3,0). x=2p+q​⇒x=27+3​=210​=5​ We found that the axis of symmetry is the vertical line x=5.Find and Plot the Vertex Since the vertex lies on the axis of symmetry, its x-coordinate is 5. To find the y-coordinate, we will substitute 5 for x in the given equation. h(x)=-4(x−7)(x−3)x=5h(5)=-4(5−7)(5−3) Simplify right-hand side Subtract termsh(5)=-4(-2)(2)-a(-b)=a⋅bh(5)=8(2)Multiply h(5)=16 The y-coordinate of the vertex is 16. Therefore, the vertex is the point (5,16).Draw the Parabola Finally, we will draw the parabola as a curve passing through the vertex and the x-intercepts.We can see that there are no restrictions on the x-variable. Furthermore, h(x) — or y — takes values less than or equal to 16. We can write the domain and range of the function using this information. Domain:Range:​ All real numbers y≤16​
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Exercises 19 To draw the graph of the given function, we will follow five steps.Rewrite the quadratic function in intercept form. Identify and plot the x-intercepts. Find and graph the axis of symmetry. Find and plot the vertex. Draw the parabola through the vertex and the points where the x-intercepts occur.Let's go through these steps one at a time.Rewrite the Function We will start by rewriting the function in intercept form. To do so, we will factor the right-hand side of the given equation. y=4x2−36x+32Factor out 4y=4(x2−9x+8)Write as a differencey=4(x2−x−8x+8) Factor out x & -8 Factor out xy=4(x(x−1)−8x+8)Factor out -8 y=4(x(x−1)−8(x−1))Factor out (x−1)y=4(x−1)(x−8)Identify and Plot the x-intercepts Recall the intercept form of a quadratic function. f(x)=a(x−p)(x−q)​ In this form, where a ≠ 0, the x-intercepts are p and q. Let's consider the intercept form of our function. h(x)=4(x−1)(x−8)​ We can see that a=4, p=1, and q=8. Therefore, the x-intercepts occur at (1,0) and (8,0).Find and Graph the Axis of Symmetry The axis of symmetry is halfway between (p,0) and (q,0). Since we know that p=1 and q=8, the axis of symmetry of our parabola is halfway between (1,0) and (8,0). x=2p+q​⇒x=21+8​=29​=4.5​ We found that the axis of symmetry is the vertical line x=4.5.Find and Plot the Vertex Since the vertex lies on the axis of symmetry, its x-coordinate is 4.5. To find the y-coordinate, we will substitute 4.5 for x in the given equation. y=4x2−36x+32x=4.5y=4(4.5)2−36(4.5)+32 Simplify right-hand side Calculate powery=4(20.25)−36(4.5)+32Multiplyy=81−36(4.5)+32(-a)b=-aby=81−162+32Add and subtract terms y=-49 The y-coordinate of the vertex is -49. Therefore, the vertex is the point (4.5,-49).Draw the Parabola Finally, we will draw the parabola through the vertex and the x-intercepts.We can see above that there are no restrictions on the x-variable. Furthermore, the y-variable takes values greater than or equal to -49. We can write the domain and range of the function using this information. Domain:Range:​ all real numbers y≥-49​
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Exercises 43 Let's start by finding the error. The student claims that the zeros of the function y=5(x+3)(x−2) are 3 and -2. To show that this is incorrect, let's substitute the zeros into the equation and see if they make y=0. y=5(x+3)(x−2)x=3y=5(3+3)(3−2) Simplify right-hand side Add and subtract termsy=5(6)(1)Multiply y=30 × We can see that x=3 is not a zero. Let's try x=-2. y=5(x+3)(x−2)x=-2y=5(-2+3)(-2−2) Simplify right-hand side Add and subtract termsy=5(1)(-4)Multiply y=-20 × This shows that x=-2 is also not a root. The error was made because the Zero-Product Property was not used. To find the zeros of a function, we want to set each factor equal to zero and solve for x. Let's correctly find the zeros by solving 0=5(x+3)(x−2). We will use the Zero-Product Property. 0=5(x+3)(x−2) Solve using the Zero Product Property Use the Zero Product Property{0=5(x+3)0=x−2​(I)(II)​(I):  LHS/5=RHS/5{0=x+30=x−2​(I)(II)​(I):  LHS−3=RHS−3{-3=x0=x−2​(I)(II)​(II):  LHS+2=RHS+2{-3=x2=x​(I)(II)​(I), (II):  Rearrange equation {x=-3x=2​(I)(II)​ This tells us that the zeros of the function are at x=-3 and x=2.
Exercises 44 Let's start by finding the error. The claim is that the zeros of the function y=(x+4)(x2−9) are -4 and 9. To show that this is incorrect, let's substitute these zeros into the equation and see if they make y=0. y=(x+4)(x2−9)x=-4y=(-4+4)(-42−9) Simplify right-hand side Add termsy=(0)(-42−9)Zero Property of Multiplication y=0 ✓ We can see that x=-4 is a zero. This is correct. Let's try x=9. y=(x+4)(x2−9)x=9y=(9+4)(92−9) Simplify right-hand side Calculate powery=(9+4)(81−9)Add and subtract termsy=(13)(72)Multiply y=936 × This shows that x=9 is not a zero. The error was made here because the Zero-Product Property was not used. To find the zeros of a function, we want to set each factor equal to zero and solve for x. Let's correctly find the zeros by solving 0=(x+4)(x2−9). We will use the Zero-Product Property. 0=(x+4)(x2−9) Solve using the Zero Product Property Use the Zero Product Property{0=x+40=x2−9​(I)(II)​(I):  LHS−4=RHS−4{-4=x0=x2−9​(I)(II)​(II):  LHS+9=RHS+9{-4=x9=x2​(I)(II)​(II):  LHS​=RHS​{-4=x±3=x​(I)(II)​(I), (II):  Rearrange equation {x=-4x=±3​(I)(II)​ This tells us that the zeros of the function are at x=-4, x=-3, and x=3.
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