$$ \left. \begin{array} { l } { h ( x + a ) = ( x + a ) ^ { 2 } - 4 ( a + 8 ) - 6 } \ { x ightarrow x + c = x ^ { 2 } + a ^ { 2 } - 4 a - 32 - 6 } \end{array} ight. $$

Question

$$ \left. \begin{array} { l } { h ( x + a ) = ( x + a ) ^ { 2 } - 4 ( a + 8 ) - 6 } \ { x ightarrow x + c = x ^ { 2 } + a ^ { 2 } - 4 a - 32 - 6 } \end{array} ight. $$

UpStudy AI · Accepted Answer

We begin with the two given relations:
$$
h(x+a) = (x+a)^2 - 4(a+8) - 6
$$
and
$$
h(x+c) = x^2 + a^2 - 4a - 32 - 6.
$$

**Step 1. Expand the first equation.**

Write
$$
(x+a)^2 = x^2 + 2ax + a^2.
$$
Then
$$
h(x+a) = x^2 + 2ax + a^2 - 4(a+8) - 6.
$$
Simplify the constant term:
$$
-4(a+8) - 6 = -4a - 32 - 6 = -4a - 38.
$$
Thus,
$$
h(x+a) = x^2 + 2ax + a^2 - 4a - 38.
$$

**Step 2. Write the second equation in simplified form.**

In the second relation, combine the constants:
$$
-32 - 6 = -38,
$$
so that
$$
h(x+c) = x^2 + a^2 - 4a - 38.
$$

**Step 3. Compare the two expressions.**

We have:
$$
h(x+a) = x^2 + 2ax + a^2 - 4a - 38
$$
and
$$
h(x+c) = x^2 + a^2 - 4a - 38.
$$
Notice that the only difference is in the coefficient of $$x$$. In $$h(x+a)$$ the coefficient of $$x$$ is $$2a$$ while in $$h(x+c)$$ the coefficient of $$x$$ is $$0$$.

Since the two expressions for $$h$$ must be the same for all $$x$$, we require that:
$$
2a = 0.
$$
This gives
$$
a = 0.
$$

**Step 4. Find the shift $$c$$.**

Now, express the function $$h$$ in general. Since in the first relation we have
$$
h(x+a) = (x+a)^2 - 4(a+8) - 6,
$$
and we have found $$a = 0$$, it becomes
$$
h(x) = x^2 - 4(0+8) - 6 = x^2 - 32 - 6 = x^2 - 38.
$$

Then, writing
$$
h(x+c) = (x+c)^2 - 38,
$$
we compare this with the given second expression:
$$
h(x+c) = x^2 + a^2 - 4a - 38.
$$
Substitute $$a = 0$$:
$$
h(x+c) = x^2 + 0^2 - 4(0) - 38 = x^2 - 38.
$$
Now expand $$(x+c)^2 - 38$$:
$$
(x+c)^2 - 38 = x^2 + 2cx + c^2 - 38.
$$
For the two forms to be identical for every $$x$$ we must have:
- Coefficient of $$x$$: $$2c = 0$$ so $$c = 0$$.
- Constant term: $$c^2 - 38 = -38$$ which is automatically true when $$c = 0$$.

**Conclusion.**

The values of the parameters are:
$$
a = 0 \quad \text{and} \quad c = 0.
$$
$$ a = 0 $$ and $$ c = 0 $$.

\( \left. \begin{array} { l } { h ( x + a ) = ( x + a ) ^ { 2 } - 4 ( a + 8 ) - 6 } \\ { x \rightarrow x + c = x ^ { 2 } + a ^ { 2 } - 4 a - 32 - 6 } \end{array} \right. \)

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Algebra Questions