Answer the questions about the following function.

(a) Is the point on the graph of ?
(b) If , what is ? What point is on the graph of ?
© If , what is ? What point(s) are on the graph of ?
(d) What is the domain of ?
(e) List the -intercept(s), if any, of the graph of .
(f) List the -intercept, if there is one, of the graph of .

Ask by Pollard Mcguire. in the United States

Mar 15,2025

Question

Answer the questions about the following function.

(a) Is the point on the graph of ?
(b) If , what is ? What point is on the graph of ?
© If , what is ? What point(s) are on the graph of ?
(d) What is the domain of ?
(e) List the -intercept(s), if any, of the graph of .
(f) List the -intercept, if there is one, of the graph of .

Ask by Pollard Mcguire. in the United States

Mar 15,2025

UpStudy AI · Accepted Answer

**(a)**  
To check if the point $$(-2,15)$$ is on the graph, substitute $$ x = -2 $$ into $$ f(x) $$:

$$
f(-2)=4(-2)^2-(-2)-3 = 4(4)+2-3=16+2-3=15.
$$

Since $$ f(-2)=15 $$, the point $$(-2,15)$$ is on the graph.

**(b)**  
Substitute $$ x=2 $$ into $$ f(x) $$:

$$
f(2)=4(2)^2-2-3 =4(4)-2-3=16-2-3=11.
$$

Thus, when $$ x=2 $$, $$ f(2)=11 $$ and the point $$(2,11)$$ is on the graph.

**(c)**  
To find $$ x $$ when $$ f(x)=-3 $$, solve:

$$
4x^2-x-3=-3.
$$

Add $$ 3 $$ to both sides:

$$
4x^2-x=0.
$$

Factor out $$ x $$:

$$
x(4x-1)=0.
$$

Setting each factor to zero gives:

$$
x=0 \quad \text{or} \quad 4x-1=0 \Rightarrow x=\frac{1}{4}.
$$

The points on the graph are $$(0,-3)$$ and $$\left(\frac{1}{4},-3\right)$$.

**(d)**  
The function $$ f(x)=4x^2-x-3 $$ is a polynomial. Therefore, its domain is all real numbers:

$$
\text{Domain: } (-\infty,\infty).
$$

**(e)**  
The $$ x $$-intercepts occur when $$ f(x)=0 $$:

$$
4x^2-x-3=0.
$$

Using the quadratic formula

$$
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}, \quad \text{with } a=4, \, b=-1, \, c=-3,
$$

compute the discriminant:

$$
\Delta=(-1)^2-4(4)(-3)=1+48=49.
$$

Thus,

$$
x=\frac{-(-1) \pm \sqrt{49}}{2(4)}=\frac{1 \pm 7}{8}.
$$

The solutions are:

$$
x=\frac{1+7}{8}=1 \quad \text{and} \quad x=\frac{1-7}{8}=-\frac{6}{8}=-\frac{3}{4}.
$$

So, the $$ x $$-intercepts are $$(1,0)$$ and $$\left(-\frac{3}{4},0\right)$$.

**(f)**  
The $$ y $$-intercept is found by evaluating $$ f(0) $$:

$$
f(0)=4(0)^2-0-3=-3.
$$

Thus, the $$ y $$-intercept is $$(0,-3)$$.
**(a)** The point $$(-2,15)$$ is on the graph of $$ f $$. **(b)** When $$ x=2 $$, $$ f(2)=11 $$, so the point $$(2,11)$$ is on the graph. **(c)** When $$ f(x)=-3 $$, $$ x=0 $$ or $$ x=\frac{1}{4} $$, so the points $$(0,-3)$$ and $$\left(\frac{1}{4},-3\right)$$ are on the graph. **(d)** The domain of $$ f $$ is all real numbers. **(e)** The $$ x $$-intercepts are $$(1,0)$$ and $$\left(-\frac{3}{4},0\right)$$. **(f)** The $$ y $$-intercept is $$(0,-3)$$.

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