Q. Determine the y-intercept of the following equation: \( (-4x-4)(3x-15) = y \).

Q: What is the y-intercept of (-4x-4)(3x-15)=y?.

Set x=0: y=(-4)(-15)=60. The y-intercept is (0,60).

Q: How do you find a y-intercept in general?

Set x=0 and evaluate y. For y=ax^2+bx+c the y-intercept equals the constant term c.

Q: What are the x-intercepts of y= (-4x-4)(3x-15)?

Set y=0. Solve (-4x-4)(3x-15)=0 giving x=-1 and x=5. Intercepts: (-1,0) and (5,0).

Q: Is this a quadratic function and what is its expanded form?

Yes. FOIL: (-4x-4)(3x-15)=-12x^2+48x+60, a quadratic with leading coefficient -12..

Q: How do you expand (-4x-4)(3x-15) ?

FOIL: (-4x)(3x)=-12x^2, (-4x)(-15)=60x, (-4)(3x)=-12x, (-4)(-15)=60. Sum: -12x^2+48x+60..

Q: \text{Which way does the parabola open and what is its maximum?}

Leading coefficient a=-12<0, so it opens downward. Maximum value is the vertex y-coordinate 108 at x=2.

Q: What are the domain and range of the function?

Domain: all real numbers. Range: (-\infty,108], since the parabola opens downward with maximum 108.

Answer

Plug in x = 0: \(y = (-4\cdot0 – 4)(3\cdot0 – 15) = (-4)(-15) = 60\).
Y-intercept: \((0, 60)\)

Detailed Explanation

Determine the y-intercept of the equation

We are given the equation

\(y = (-4x – 4)(3x – 15)\).

Step-by-step solution:

Recall the definition of the y-intercept:The y-intercept is the point where the graph crosses the y-axis. That occurs when the x-coordinate is 0. So to find the y-intercept we evaluate y at x = 0.
Substitute x = 0 into the equation:Replace x by 0 in the expression for y:
\(y = (-4(0) – 4)(3(0) – 15)\).
Simplify each factor:Compute each parenthesis separately:
\(-4(0) – 4 = 0 – 4 = -4\)

\(3(0) – 15 = 0 – 15 = -15\)
Multiply the results:Now multiply the two numbers:
\(y = (-4)(-15) = 60\).
State the y-intercept as a point:The y-intercept is the point with x = 0 and y = 60, so the y-intercept is
\((0, 60)\).

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Algebra FAQs

What is the y-intercept of \((-4x-4)(3x-15)=y\)?.

Set \(x=0\): \(y=(-4)(-15)=60\). The y-intercept is \((0,60)\).

How do you find a y-intercept in general?

Set \(x=0\) and evaluate \(y\). For \(y=ax^2+bx+c\) the \(y\)-intercept equals the constant term \(c\).

What are the x-intercepts of \(y= (-4x-4)(3x-15)\)?

Set \(y=0\). Solve \((-4x-4)(3x-15)=0\) giving \(x=-1\) and \(x=5\). Intercepts: \((-1,0)\) and \((5,0)\).

Is this a quadratic function and what is its expanded form?

Yes. FOIL: \((-4x-4)(3x-15)=-12x^2+48x+60\), a quadratic with leading coefficient \(-12\)..

How do you expand \( (-4x-4)(3x-15) \)?

FOIL: \((-4x)(3x)=-12x^2\), \((-4x)(-15)=60x\), \((-4)(3x)=-12x\), \((-4)(-15)=60\). Sum: \(-12x^2+48x+60\)..

What is the vertex and axis of symmetry?

\( \text{Which way does the parabola open and what is its maximum?} \)

Leading coefficient \(a=-12<0\), so it opens downward. Maximum value is the vertex y-coordinate \(108\) at \(x=2\).

What are the domain and range of the function?

Domain: all real numbers. Range: \((-\infty,108]\), since the parabola opens downward with maximum \(108\).

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